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Electric Flux & Gauss's Law

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Q1MHT CET 2026 · 16 APR · SHIFT 1Easy

Electric Flux & Gauss's Law

The electric field in the region is E⃗=ai^+bj^\vec{E} = a\hat{i} + b\hat{j} where 'a' and 'b' are constants. The net electric flux passing through a square area of side 'l' parallel to Y-Z plane is

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Answer: (d)

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Q2MHT CET 2026 · 15 APR · SHIFT 1Easy

Electric Flux & Gauss's Law

The number of lines of force originating from a point charge of 2×8.85×10−92 \times 8.85 \times 10^{-9} C in a medium of dielectric constant 1010 is (ϵ0=8.85×10−12\epsilon_0 = 8.85 \times 10^{-12} SI unit)

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Answer: (b)

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Q3MHT CET 2026 · 14 MAY · SHIFT 1Easy

Electric Flux & Gauss's Law

The electric flux entering and leaving an enclosed surface are Φ1\Phi_1 and Φ2\Phi_2 respectively. The electric charge inside the surface will be (ε0\varepsilon_0 = permittivity of vacuum)

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Answer: (d)

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Q4MHT CET 2026 · 13 APR · SHIFT 1Easy

Electric Flux & Gauss's Law

A charge 'Q' C is placed at the centre of a cube. The electric flux through two opposite faces of the cube is

(ϵ0\epsilon_0 = permittivity of free space)

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Answer: (b)

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Q5MHT CET 2026 · 11 APR · SHIFT 2Easy

Electric Flux & Gauss's Law

If the radius of the spherical Gaussian surface is increased then the electric flux due to a point charge enclosed by the surface

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Answer: (b)

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Q6JEE Main 2026 · 2 APR · SHIFT 2Easy

Electric Flux & Gauss's Law

Two point charges 8 μC8\ \mu\text{C} and −2 μC-2\ \mu\text{C} are located at x=2x = 2 cm and x=4x = 4 cm, respectively on the xx-axis. The ratio of electric flux due to these charges through two spheres of radii 3 cm and 5 cm with their centers at the origin is ________.

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Answer: (c)

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Q7MHT CET 2026 · 14 MAY · SHIFT 2Easy

Electric Flux & Gauss's Law

An infinite line charge produces a field of 9×1089 \times 10^{8} N/C at a distance of 22 cm. The linear charge density is

(14πε0=9×109\left(\frac{1}{4\pi\varepsilon_0} = 9 \times 10^{9}\right. SI unit)\left.\right)

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Answer: (a)

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Q8MHT CET 2026 · 15 APR · SHIFT 2Easy

Electric Flux & Gauss's Law

For a uniformly charged plane sheet, the variation of electric field (E) with distance (d) is correctly shown graphically in graph

Figure

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Answer: (c)

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Q9MHT CET 2026 · 14 MAY · SHIFT 2Easy

Electric Flux & Gauss's Law

The electric field intensity at a point near and outside the surface of a charged conductor of any shape is E1E_1. The electric field intensity due to a uniformly charged infinite thin plane sheet having the same surface charge density is E2E_2. The relation between E1E_1 and E2E_2 is

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Answer: (c)

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Q10NEET 2026Easy

Electric Flux & Gauss's Law

Which of the following statements are correct?

A. Inside a conductor, the electrostatic field is zero.

B. Electric field at the surface of a charged conductor does not depend on its surface charge density.

C. The interior of a charged conductor can have no excess charge in the static situation. D. At the surface of a charged conductor, the electrostatic field must be normal to the surface at every point.

E. The electrostatic potential is zero everywhere inside a charged conductor.

Choose the correct answer from the options given below:

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Answer: (c)

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Q11JEE Main 2026 · 4 APR · SHIFT 2Easy

Electric Flux & Gauss's Law

Given below are two statements: one is labelled as Assertion A and the other is labelled as Reason R\mathbf{R}

Assertion A: In electrostatics, a conductor does not store any net charge inside. Reason R: Inside the capacitor (with no dielectric medium), the free charge carriers, if placed between the plates of capacitor, experience force and drift.

Choose the correct answer from the options given below

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Answer: (b)

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Q12JEE Main 2026 · 23 JAN · SHIFT 1Medium

Electric Flux & Gauss's Law

Two point charges 2q2q and qq are placed at vertex AA and centre of face CDEFCDEF of the cube as shown in figure. The electric flux passing through the cube is:

Figure

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Answer: (b)

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Q13MHT CET 2026 · 12 MAY · SHIFT 2Medium

Electric Flux & Gauss's Law

A spherical conductor of diameter 33 mm is kept in a uniform electric field of intensity 4×1074 \times 10^{7} N/C. The maximum charge on the conductor is (14πϵ0=9×109 SI units)\left(\frac{1}{4\pi\epsilon_0} = 9 \times 10^{9}\text{ SI units}\right)

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Answer: (b)

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Q14NEET 2026 · RE-NEETMedium

Electric Flux & Gauss's Law

A point charge QQ is placed inside a cavity within a solid isolated conducting sphere. Consider points A,BA, B and CC as shown in the figure, where the magnitudes of the electric fields are EA,EB,ECE_A, E_B, E_C, respectively. The points BB and CC are at the same distance from the center of the solid sphere. The correct option is:

Figure

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Answer: (c)

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Q15JEE Main 2025 · 23 JAN · SHIFT 1Easy

Electric Flux & Gauss's Law

The electric flux is ϕ=ασ+βλ\phi = \alpha\sigma + \beta\lambda where λ\lambda and σ\sigma are linear and surface charge density, respectively. (αβ)\left(\frac{\alpha}{\beta}\right) represents

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Answer: (a)

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Q16JEE Main 2025 · 7 APR · SHIFT 2Easy

Electric Flux & Gauss's Law

The electric field in a region is given by E⃗=(2i^+4j^+6k^)×103 N/C\vec{E} = (2\hat{i} + 4\hat{j} + 6\hat{k}) \times 10^{3}\text{ N/C}. The flux of the field through a rectangular surface parallel to x−zx - z plane is 6.0 N m2C−16.0\text{ N m}^2\text{C}^{-1}. The area of the surface is _____________ cm2\text{cm}^2.

Numerical answer type. Work it out, then open the answer.

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Answer: 15

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Q17MHT CET 2025 · 5 MAY · SHIFT 2Easy

Electric Flux & Gauss's Law

The electric flux through the surface

Figure

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Answer: (c)

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Q18MHT CET 2025 · 5 MAY · SHIFT 2Easy

Electric Flux & Gauss's Law

A charge is uniformly distributed on the surface of a spherical rubber balloon. As it is blown up, the total electric flux coming out of the surface

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Answer: (c)

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Q19MHT CET 2025 · 22 APR · SHIFT 2Easy

Electric Flux & Gauss's Law

A charge Q μCQ\ \mu\text{C} is placed at the centre of a cube. The flux through two opposite faces of the cube is ( ε0=\varepsilon_0 = permittivity of free space)

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Answer: (b)

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Q20MHT CET 2025 · 22 APR · SHIFT 1Easy

Electric Flux & Gauss's Law

Four charges 2 μC, −3 μC, 4 μC, −4 μC2\ \mu\text{C},\ -3\ \mu\text{C},\ 4\ \mu\text{C},\ -4\ \mu\text{C} and −1 μC-1\ \mu\text{C} are enclosed by the Gaussian surface of radius 2 m. Net outward flux through the Gaussian surface is (in μV-m\mu\text{V-m}) [ ε0=\varepsilon_0 = permittivity of free space]

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Answer: (a)

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Q21MHT CET 2025 · 20 APR · SHIFT 2Easy

Electric Flux & Gauss's Law

A hollow cylinder has a charge ' qq ' coulomb within it. If ' ϕ\phi ' is the electric flux in unit of V-m associated with the curved surface C, the flux linked with the plane surface ' AA ' in unit of V-m will be [ ε0=\varepsilon_0 = permittivity of free space]0

Figure

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Answer: (c)

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Q22MHT CET 2025 · 20 APR · SHIFT 1Easy

Electric Flux & Gauss's Law

A hollow cylinder has a charge of ' qq ' CC within it. If ϕ\phi is the electric flux associated with the curved surface B, the flux linked with the plane surface A will be

Figure

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Answer: (d)

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Q23JEE Main 2025 · 29 JAN · SHIFT 2Easy

Electric Flux & Gauss's Law

A point charge causes an electric flux of −2×104 N m2C−1-2 \times 10^{4}\text{ N m}^2\text{C}^{-1} to pass through a spherical Gaussian surface of 8.0 cm radius, centred on the charge. The value of the point charge is: (Given ϵ0=8.85×10−12 C2N−1m−2\epsilon_0 = 8.85 \times 10^{-12}\text{ C}^2\text{N}^{-1}\text{m}^{-2})

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Answer: (b)

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Q24MHT CET 2025 · 21 APR · SHIFT 2Easy

Electric Flux & Gauss's Law

The electric field intensity on the surface of a solid charged sphere of radius r\mathbf{r} and volume charge density σ\sigma is ( ε0=\varepsilon_0 = permittivity of free space)

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Answer: (d)

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Q25MHT CET 2025 · 23 APR · SHIFT 1Easy

Electric Flux & Gauss's Law

The electric field intensity near a conducting surface having uniform surface charge density ' σ\sigma ' is given by ( ε0=\varepsilon_0 = permittivity of free space)

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Answer: (c)

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Q26JEE Main 2025 · 8 APR · SHIFT 2Easy

Electric Flux & Gauss's Law

Electric charge is transferred to an irregular metallic disk as shown in the figure. If σ1,σ2,σ3\sigma_1, \sigma_2, \sigma_3 and σ4\sigma_4 are charge densities at given points then, choose the correct answer from the options given below:

Figure

A. σ1>σ3;σ2=σ4\sigma_1 > \sigma_3; \sigma_2 = \sigma_4

B. σ1>σ2;σ3>σ4\sigma_1 > \sigma_2; \sigma_3 > \sigma_4

C. σ1>σ3>σ2=σ4\sigma_1 > \sigma_3 > \sigma_2 = \sigma_4

D. σ1<σ3<σ2=σ4\sigma_1 < \sigma_3 < \sigma_2 = \sigma_4

E. σ1=σ2=σ3=σ4\sigma_1 = \sigma_2 = \sigma_3 = \sigma_4

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Answer: (d)

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Q27JEE Main 2025 · 7 APR · SHIFT 2Easy

Electric Flux & Gauss's Law

Given below are two statements: one is labelled as Assertion (A) and the other is labelled as Reason (R).

Assertion (A): The outer body of an aircraft is made of metal which protects persons sitting inside from lightning strikes.

Reason (R): The electric field inside the cavity enclosed by a conductor is zero. In the light of the above statements, choose the most−-appropriate answer from the options given below:

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Answer: (c)

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Q28JEE Main 2025 · 2 APR · SHIFT 1Easy

Electric Flux & Gauss's Law

Consider two infinitely large plane parallel conducting plates as shown below. The plates are uniformly charged with a surface charge density +σ+\sigma and −2σ-2\sigma. The force experienced by a point charge +q placed at the mid point between two plates will be:

Figure

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Answer: (b)

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Q29MHT CET 2025 · 19 APR · SHIFT 2Medium

Electric Flux & Gauss's Law

A uniformly charged conducting sphere of diameter 3.5 cm has a surface charge density of 20 μC m−220\ \mu\text{C m}^{-2}. The total electric flux leaving the surface of the sphere is nearly [permittivity of free space, ε0=8.85×10−12 SI\varepsilon_0 = 8.85 \times 10^{-12}\text{ SI} unit]

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Answer: (c)

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Q30JEE Main 2025 · 8 APR · SHIFT 2Medium

Electric Flux & Gauss's Law

An infinitely long wire has uniform linear charge density λ=2\lambda = 2 nC/m. The net flux through a Gaussian cube of side length 3\sqrt{3} cm, if the wire passes through any two corners of the cube, that are maximally displaced from each other, would be xx Nm2C−1{}^{2}\text{C}^{-1}, where xx is:

[Neglect any edge effects and use 14πϵ0=9×109\frac{1}{4\pi\epsilon_0} = 9 \times 10^{9} SI units]

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Answer: (d)

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Q31JEE Main 2025 · 22 JAN · SHIFT 1Medium

Electric Flux & Gauss's Law

A line charge of length a2\frac{a}{2} is kept at the center of an edge BCBC of a cube ABCDEFGH having edge length ' aa ' as shown in the figure. If the density of line charge is λ C\lambda\text{ C} per unit length, then the total electric flux through all the faces of the cube will be ___________. (Take, ϵ0\epsilon_0 as the free space permittivity)

Figure

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Answer: (b)

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Q32JEE Main 2025 · 24 JAN · SHIFT 1Medium

Electric Flux & Gauss's Law

A square loop of sides a=1 ma = 1\text{ m} is held normally in front of a point charge q=1 Cq = 1\text{ C} at a distance a2\frac{a}{2}. The flux of the electric field through the shaded region is 5p×1ε0 Nm2C\frac{5}{p} \times \frac{1}{\varepsilon_0}\,\frac{\text{Nm}^{2}}{\text{C}}, where the value of p is ________.

Figure

Numerical answer type. Work it out, then open the answer.

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Answer: 48

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Q33JEE Main 2025 · 4 APR · SHIFT 1Medium

Electric Flux & Gauss's Law

Two infinite identical charged sheets and a charged spherical body of charge density ' ρ\rho ' are arranged as shown in figure. Then the correct relation between the electrical fields at A,B,CA, B, C and D points is:

Figure

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Answer: (c)

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Q34JEE Main 2025 · 29 JAN · SHIFT 1Medium

Electric Flux & Gauss's Law

Match List - I with List - II.

Choose the correct answer from the options given below:

List - IList - II
(A) Electric field inside (distance r > 0 from center) of a uniformly charged spherical shell with surface charge density σ, and radius R.(I) σ/ε0
(B) Electric field at distance r>0 from a uniformly charged infinite plane sheet with surface charge density σ.(II) σ/2ε0
(C) Electric field outside (distance r>0 from center) of a uniformly charged spherical shell with surface charge density σ, and radius R.(III) 0
(D) Electric field between 2 oppositely charged infinite plane parallel sheets with uniform surface charge density σ.(IV) σϵ0r2\frac{\sigma}{\epsilon_0 r^{2}}
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Answer: (c)

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Q35JEE Main 2024 · 29 JAN · SHIFT 2Easy

Electric Flux & Gauss's Law

An electric field is given by (6i^+5j^+3k^) N/C(6\hat{i} + 5\hat{j} + 3\hat{k})\text{ N/C}. The electric flux through a surface area 30i^ m230\hat{i}\text{ m}^2 lying in YZ-plane (in SI unit) is:

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Answer: (c)

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Q36JEE Main 2024 · 8 APR · SHIFT 1Easy

Electric Flux & Gauss's Law

An electric field, E⃗=2i^+6j^+8k^6\vec{E} = \frac{2\hat{i} + 6\hat{j} + 8\hat{k}}{\sqrt{6}} passes through the surface of 4 m24\text{ m}^2 area having unit vector n^=(2i^+j^+k^6)\hat{n} = \left(\frac{2\hat{i} + \hat{j} + \hat{k}}{\sqrt{6}}\right). The electric flux for that surface is _________ V m\text{V m}.

Numerical answer type. Work it out, then open the answer.

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Answer: 12

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Q37MHT CET 2024 · 16 MAY · SHIFT 2Easy

Electric Flux & Gauss's Law

A hollow cylinder has charge ' qq ' CC within it. If ' ϕ\phi ' is the electric flux associated with the curved surface B, the flux linked with the plane surface A will be:

Figure

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Answer: (a)

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Q38MHT CET 2024 · 9 MAY · SHIFT 2Easy

Electric Flux & Gauss's Law

The electric flux over a sphere of radius ' rr ' is ' ϕ\phi '. If the radius of the sphere is doubled without changing the charge, the flux will be

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Answer: (c)

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Q39MHT CET 2024 · 9 MAY · SHIFT 1Easy

Electric Flux & Gauss's Law

If the electric flux entering and leaving an enclosed surface is ϕ1\phi_1 and ϕ2\phi_2 then charge enclosed in the surface is (ε0=\varepsilon_0 = permittivity of free space)

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Answer: (d)

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Q40MHT CET 2024 · 4 MAY · SHIFT 2Easy

Electric Flux & Gauss's Law

Two surfaces A and B are enclosing the charges as shown below. The total normal electric induction (T.N.E.I) through the surfaces A and B are respectively.

Figure

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Answer: (c)

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Q41MHT CET 2024 · 4 MAY · SHIFT 2Easy

Electric Flux & Gauss's Law

A spherical rubber balloon carries a charge, uniformly distributed over the surface. As the balloon is blown up and increases in size, the total electric flux coming out of the surface

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Answer: (d)

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Q42MHT CET 2024 · 4 MAY · SHIFT 2Easy

Electric Flux & Gauss's Law

If the electric flux entering and leaving an enclosed surface are ϕ1\phi_1 and ϕ2\phi_2 respectively, the electric charge inside the surface will be

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Answer: (d)

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Q43JEE Main 2024 · 9 APR · SHIFT 2Easy

Electric Flux & Gauss's Law

Five charges +q,+5q,−2q,+3q+q, +5q, -2q, +3q and −4q-4q are situated as shown in the figure. The electric flux due to this configuration through the surface SS is:

Figure

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Answer: (d)

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Q44JEE Main 2024 · 4 APR · SHIFT 2Easy

Electric Flux & Gauss's Law

A charge qq is placed at the center of one of the surfaces of a cube. The flux linked with the cube is:

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Answer: (a)

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Q45JEE Main 2024 · 1 FEB · SHIFT 2Easy

Electric Flux & Gauss's Law

C1C_1 and C2C_2 are two hollow concentric cubes enclosing charges 2Q2Q and 3Q3Q respectively as shown in figure. The ratio of electric flux passing through C1C_1 and C2C_2 is:

Figure

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Answer: (c)

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Q46JEE Main 2024 · 29 JAN · SHIFT 1Easy

Electric Flux & Gauss's Law

Two charges of 5Q5Q and −2Q-2Q are situated at the points (3a,0)(3a, 0) and (−5a,0)(-5a, 0) respectively. The electric flux through a sphere of radius '4a4a' having center at origin is:

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Answer: (d)

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Q47JEE Main 2024 · 6 APR · SHIFT 1Easy

Electric Flux & Gauss's Law

σ\sigma is the uniform surface charge density of a thin spherical shell of radius R. The electric field at any point on the surface of the spherical shell is:

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Answer: (b)

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Q48MHT CET 2024 · 9 MAY · SHIFT 2Easy

Electric Flux & Gauss's Law

The van de Graaff Generator is not based on

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Answer: (b)

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Q49JEE Main 2024 · 5 APR · SHIFT 2Easy

Electric Flux & Gauss's Law

The vehicles carrying inflammable fluids usually have metallic chains touching the ground:

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Answer: (d)

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Q50JEE Main 2024 · 9 APR · SHIFT 2Medium

Electric Flux & Gauss's Law

An electric field E⃗=(2xi^) NC−1\vec{E} = (2x\hat{i})\text{ NC}^{-1} exists in space. A cube of side 2 m2\text{ m} is placed in the space as per figure given below. The electric flux through the cube is ______ Nm2/C\text{Nm}^{2}/\text{C}.

Figure

Numerical answer type. Work it out, then open the answer.

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Answer: 16

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Q51MHT CET 2024 · 2 MAY · SHIFT 1Medium

Electric Flux & Gauss's Law

A uniformly charged conducting sphere of diameter 14 cm has surface charge density of 40 μC m−240\ \mu\text{C m}^{-2}. The total electric flux leaving the surface of the sphere is nearly (Permittivity of free space =8.85×10−12= 8.85 \times 10^{-12} SI unit)

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Answer: (d)

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Q52MHT CET 2024 · 15 MAY · SHIFT 1Medium

Electric Flux & Gauss's Law

Consider a long uniformly charged cylinder having constant volume charge density ' λ\lambda ' and radius ' RR '. A Gaussian surface is in the form of a cylinder of radius ' rr ' such that vertical axis of both the cylinders coincide. For a point inside the cylinder (r<R)(r < R), electric field is directly proportional to

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Answer: (b)

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Q53JEE Main 2024 · 6 APR · SHIFT 1Medium

Electric Flux & Gauss's Law

Three infinitely long charged thin sheets are placed as shown in figure. The magnitude of electric field at the point PP is xσϵ0\frac{x\sigma}{\epsilon_0}. The value of xx is _________ (all quantities are measured in SI units).

Figure

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Answer: 2

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Q54JEE Main 2024 · 4 APR · SHIFT 1Medium

Electric Flux & Gauss's Law

An infinite plane sheet of charge having uniform surface charge density +σs C/m2+\sigma_s\text{ C/m}^{2} is placed on x-yx\text{-}y plane. Another infinitely long line charge having uniform linear charge density +λe C/m+\lambda_e\text{ C/m} is placed at z=4 mz = 4\text{ m} plane and parallel to yy-axis. If the magnitude values ∣σs∣=2∣λe∣|\sigma_s| = 2|\lambda_e| then at point (0,0,2)(0, 0, 2), the ratio of magnitudes of electric field values due to sheet charge to that of line charge is πn:1\pi\sqrt{n} : 1. The value of nn is _________.

Numerical answer type. Work it out, then open the answer.

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Answer: 16

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Q55NEET 2023Easy

Electric Flux & Gauss's Law

According to Gauss law of electrostatics, electric flux through a closed surface depends on:

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Answer: (b)

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Q56NEET 2023Easy

Electric Flux & Gauss's Law

A charge Q μCQ\ \mu\text{C} is placed at the centre of a cube. The flux coming out from any one of its faces will be (in SI unit):

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Answer: (d)

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Q57NEET 2023Easy

Electric Flux & Gauss's Law

If ∮sE⃗⋅dS→=0\oint_s \vec{E} \cdot \overrightarrow{dS} = 0 over a surface, then:

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Answer: (d)

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Q58JEE Main 2023 · 12 APR · SHIFT 1Easy

Electric Flux & Gauss's Law

Given below are two statements: one is labelled as Assertion A and the other is labelled as Reason R

Assertion A: If an electric dipole of dipole moment 30×10−5 C m30 \times 10^{-5}\text{ C m} is enclosed by a closed surface, the net flux coming out of the surface will be zero.

Reason R: Electric dipole consists of two equal and opposite charges.

In the light of the above statements, choose the correct answer from the options given below.

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Answer: (d)

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Q59MHT CET 2023 · 14 MAY · SHIFT 2Easy

Electric Flux & Gauss's Law

A charge 17.7×10−4 C17.7 \times 10^{-4}\text{ C} is distributed uniformly over a large sheet of area 200 m2200\text{ m}^2. The electric field intensity at a distance 20 cm20\text{ cm} from it in air will be [ε0=8.85×10−12 C2/N m2]\left[\varepsilon_0 = 8.85 \times 10^{-12}\text{ C}^2/\text{N m}^2\right]

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Answer: (a)

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Q60MHT CET 2023 · 14 MAY · SHIFT 1Easy

Electric Flux & Gauss's Law

The electric field intensity on the surface of a solid charged sphere of radius 'rr' and volume charge density 'ρ\rho' is (ε0=\varepsilon_0 = permittivity of free space)

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Answer: (a)

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Q61JEE Main 2023 · 8 APR · SHIFT 1Easy

Electric Flux & Gauss's Law

Graphical variation of electric field due to a uniformly charged insulating solid sphere of radius RR, with distance rr from the centre O is represented by:

Figure

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Answer: (d)

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Q62JEE Main 2023 · 1 FEB · SHIFT 1Easy

Electric Flux & Gauss's Law

Let σ\sigma be the uniform surface charge density of two infinite thin plane sheets shown in figure. Then the electric fields in three different regions EI,EIIE_{I}, E_{II} and EIIIE_{III} are:

Figure

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Answer: (c)

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Q63MHT CET 2023 · 13 MAY · SHIFT 1Easy

Electric Flux & Gauss's Law

Which of the following statements is 'WRONG' for the conductors?

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Answer: (c)

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Q64JEE Main 2023 · 30 JAN · SHIFT 2Easy

Electric Flux & Gauss's Law

As shown in the figure, a point charge QQ is placed at the centre of conducting spherical shell of inner radius aa and outer radius bb. The electric field due to charge QQ in three different regions I, II\text{I, II} and III\text{III} is given by:

(I:r<a,II:a<r<b(\text{I}: r < a, \text{II}: a < r < b, III: r>br > b)

Figure

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Answer: (c)

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Q65JEE Main 2023 · 1 FEB · SHIFT 2Medium

Electric Flux & Gauss's Law

A cubical volume is bounded by the surfaces x=0,x=a,y=0,y=a,z=0,z=ax = 0, x = a, y = 0, y = a, z = 0, z = a. The electric field in the region is given by E⃗=E0xi^\vec{E} = E_0 x\hat{i}. Where E0=4×104 NC−1 m−1E_0 = 4 \times 10^{4}\text{ NC}^{-1}\text{ m}^{-1}. If a=2 cma = 2\text{ cm}, the charge contained in the cubical volume is Q×10−14 CQ \times 10^{-14}\text{ C}. The value of QQ is ________________. (Take ϵ0=9×10−12 C2/Nm2\epsilon_0 = 9 \times 10^{-12}\text{ C}^{2}/\text{Nm}^{2})

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Q66JEE Main 2023 · 31 JAN · SHIFT 1Medium

Electric Flux & Gauss's Law

Expression for an electric field is given by E⃗=4000x2i^ Vm\vec{E} = 4000x^{2}\hat{i}\ \frac{\text{V}}{\text{m}}. The electric flux through the cube of side 20 cm20\text{ cm} when placed in electric field (as shown in the figure) is __________ V cm\text{V cm}.

Figure

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Answer: 640

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Q67JEE Main 2023 · 29 JAN · SHIFT 1Medium

Electric Flux & Gauss's Law

In a cuboid of dimension 2L×2L×L2L \times 2L \times L, a charge qq is placed at the center of the surface 'SS' having area of 4L24L^{2}. The flux through the opposite surface to 'SS' is given by

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Answer: (d)

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Q68JEE Main 2023 · 30 JAN · SHIFT 2Medium

Electric Flux & Gauss's Law

As shown in figure, a cuboid lies in a region with electric field E=2x2i^−4yj^+6k^ N/CE = 2x^{2}\hat{i} - 4y\hat{j} + 6\hat{k}\text{ N/C}. The magnitude of charge within the cuboid is nϵ0 Cn\epsilon_0\text{ C}.

The value of nn is _________ (if dimension of cuboid is 1×2×3 m31 \times 2 \times 3\text{ m}^{3})

Figure

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Answer: 12

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Q69JEE Main 2023 · 13 APR · SHIFT 1Medium

Electric Flux & Gauss's Law

A thin infinite sheet charge and an infinite line charge of respective charge densities +σ+\sigma and +λ+\lambda are placed parallel at 5 m5\text{ m} distance from each other. Points 'P' and 'Q' are at 3π\frac{3}{\pi} m and 4π\frac{4}{\pi} m perpendicular distances from line charge towards sheet charge, respectively. 'EPE_P ' and 'EQE_Q' are the magnitudes of resultant electric field intensities at point 'P' and 'Q', respectively. If EPEQ=4a\frac{E_P}{E_Q} = \frac{4}{a} for 2∣σ∣=∣λ∣2|\sigma| = |\lambda|, then the value of aa is ___________.

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Answer: 6

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Q70MHT CET 2023 · 14 MAY · SHIFT 1Medium

Electric Flux & Gauss's Law

A conducting sphere of radius 0.1 m0.1\text{ m} has uniform charge density 1.8 μC/m21.8\ \mu\text{C/m}^{2} on its surface. The electric field in free space at radial distance 0.2 m0.2\text{ m} from a point on the surface is ( ε0=\varepsilon_0 = permittivity of free space)

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Answer: (c)

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Q71MHT CET 2023 · 9 MAY · SHIFT 1Medium

Electric Flux & Gauss's Law

A solid metallic sphere has a charge +3Q+3Q. Concentric with this sphere is a conducting spherical shell having charge −Q-Q. The radius of the sphere is 'AA' and that of the spherical shell is 'BB'. (B>A)(B > A). The electric field at a distance 'RR' (A<R<B)(A < R < B) from the centre is ( ε0=\varepsilon_0 = permittivity of vacuum)

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Q72MHT CET 2022 · 11 AUG · SHIFT 2Easy

Electric Flux & Gauss's Law

If the radius of the spherical gaussian surface is increased then the electric flux due to a point charge enclosed by the surface

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Answer: (b)

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Q73JEE Main 2022 · 25 JUN · SHIFT 1Easy

Electric Flux & Gauss's Law

In the figure, a very large plane sheet of positive charge is shown. P1 and P2 are two points at distance l and 2l from the charge distribution. If σ\sigma is the surface charge density, then the magnitude of electric fields E1 and E2 at P1 and P2 respectively are:

Figure

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Q74JEE Main 2022 · 27 JUL · SHIFT 1Easy

Electric Flux & Gauss's Law

A long cylindrical volume contains a uniformly distributed charge of density ρ C m−3\rho\text{ C m}^{-3}. The electric field inside the cylindrical volume at a distance x=2ε0ρ mx = \frac{2\varepsilon_0}{\rho}\text{ m} from its axis is ________ V m−1\text{V m}^{-1}.

Figure

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Q75JEE Main 2022 · 29 JUL · SHIFT 1Easy

Electric Flux & Gauss's Law

Given below are two statements.

Statement I: Electric potential is constant within and at the surface of each conductor. Statement II: Electric field just outside a charged conductor is perpendicular to the surface of the conductor at every point.

In the light of the above statements, choose the most appropriate answer from the options given below.

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Answer: (a)

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Q76JEE Main 2022 · 27 JUN · SHIFT 2Medium

Electric Flux & Gauss's Law

If a charge q is placed at the centre of a closed hemispherical non-conducting surface, the total flux passing through the flat surface would be:

Figure

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Answer: (b)

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Q77JEE Main 2022 · 25 JUL · SHIFT 1Medium

Electric Flux & Gauss's Law

The volume charge density of a sphere of radius 6 m6\text{ m} is 2 μC cm−32\ \mu\text{C cm}^{-3}. The number of lines of force per unit surface area coming out from the surface of the sphere is _______________ ×1010 NC−1\times 10^{10}\text{ NC}^{-1}. [Given: Permittivity of vacuum ϵ0=8.85×10−12 C2 N−1-m−2\epsilon_0 = 8.85 \times 10^{-12}\text{ C}^{2}\text{ N}^{-1}\text{-m}^{-2} ]

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Answer: 45

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Q78JEE Main 2022 · 29 JUL · SHIFT 1Difficult

Electric Flux & Gauss's Law

A spherically symmetric charge distribution is considered with charge density varying as ρ(r)=ρ0(34−rR)\rho(r) = \rho_0\left(\frac{3}{4} - \frac{r}{R}\right) for r≤Rr \leq R

zero for r>Rr > R

Where, r (r<R)r\,(r < R) is the distance from the centre O (as shown in figure). The electric field at point P will be:

Figure

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Answer: (c)

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Q79JEE Main 2021 · 17 MAR · SHIFT 2Easy

Electric Flux & Gauss's Law

The electric field in a region is given by E⃗=25E0i^+35E0j^\vec{E} = \frac{2}{5}E_0\hat{i} + \frac{3}{5}E_0\hat{j} with E0=4.0×103 NCE_0 = 4.0 \times 10^{3}\ \frac{\text{N}}{\text{C}}. The flux of this field through a rectangular surface area 0.4 m2 parallel to the Y-Z plane is __________ Nm2C−1{}^{-1}.

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Answer: 640

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Q80JEE Main 2021 · 25 FEB · SHIFT 1Easy

Electric Flux & Gauss's Law

The electric field in a region is given by E⃗=(35E0i^+45E0j^)NC\vec{E} = \left(\frac{3}{5}E_0\hat{i} + \frac{4}{5}E_0\hat{j}\right)\frac{\text{N}}{\text{C}}. The ratio of flux of reported field through the rectangular surface of area 0.2 m2 (parallel to y −- z plane) to that of the surface of area 0.3 m2 (parallel to x −- z plane) is a: b, where a = __________ [Here i^ j^\hat{i}\ \hat{j} and k^\hat{k} are unit vectors along x, y and z-axes respectively.]

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Q81JEE Main 2021 · 26 FEB · SHIFT 2Easy

Electric Flux & Gauss's Law

Given below are two statements:

Statement I: An electric dipole is placed at the center of a hollow sphere. The flux of the electric field through the sphere is zero but the electric field is not zero anywhere in the sphere. Statement II: If R is the radius of a solid metallic sphere and Q be the total charge on it. The electric field at any point on the spherical surface of radius r (< R) is zero but the electric flux passing through this closed spherical surface of radius r is not zero.

In the light of the above statements, choose the correct answer from the options given below:

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Answer: (c)

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Q82MHT CET 2021 · 24 SEP · SHIFT 1Easy

Electric Flux & Gauss's Law

The electric field intensity on the surface of a charged solid sphere of radius 'rr' and volume charge density 'ρ\rho' is given by (ϵ0=\epsilon_0 = permittivity of free space)

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Q83JEE Main 2021 · 1 SEP · SHIFT 2Medium

Electric Flux & Gauss's Law

A cube is placed inside an electric field, E⃗=150y2j^\vec{E} = 150y^{2}\hat{j}. The side of the cube is 0.5 m and is placed in the field as shown in the given figure. The charge inside the cube is:

Figure

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Answer: (b)

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Q84JEE Main 2021 · 31 AUG · SHIFT 2Medium

Electric Flux & Gauss's Law

Choose the incorrect statement:

(1) The electric lines of force entering into a Gaussian surface provide negative flux. (2) A charge 'q' is placed at the centre of a cube. The flux through all the faces will be the same. (3) In a uniform electric field net flux through a closed Gaussian surface containing no net charge, is zero.

(4) When electric field is parallel to a Gaussian surface, it provides a finite non-zero flux. Choose the most appropriate answer from the options given below

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Answer: (c)

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Q85JEE Main 2021 · 25 FEB · SHIFT 2Medium

Electric Flux & Gauss's Law

A charge 'q' is placed at one corner of a cube as shown in figure. The flux of electrostatic field E⃗\vec{E} through the shaded area is:

Figure

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Q86JEE Main 2021 · 22 JUL · SHIFT 2Medium

Electric Flux & Gauss's Law

The total charge enclosed in an incremental volume of 2 ×\times 10−-9 m3 located at the origin is ___________ nC, if electric flux density of its field is found as

D = e−-x sin y i^−\hat{i} - e−-x cos y j^\hat{j} + 2z k^\hat{k} C/m2

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Answer: 4

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Q87JEE Main 2021 · 24 FEB · SHIFT 2Medium

Electric Flux & Gauss's Law

A point charge of +12μ\muC is at a distance 6 cm vertically above the centre of a square of side 12 cm as shown in figure. The magnitude of the electric flux through the square will be __________ ×\times 103 Nm2/C.

Figure

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Answer: 226

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Q88JEE Main 2021 · 16 MAR · SHIFT 2Medium

Electric Flux & Gauss's Law

Find out the surface charge density at the intersection of point x = 3 m plane and x-axis, in the region of uniform line charge of 8 nC/m lying along the z-axis in free space.

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Q89MHT CET 2021 · 20 SEP · SHIFT 1Medium

Electric Flux & Gauss's Law

A spherical conducting shell of inner radius 'r 1{}_{1}' and outer radius 'r2{}_{2} ' has a charge 'Q'. A charge −-q is placed at the centre of the shell. The surface charge density on the inner and outer surface of the shell will be

Figure

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Q90MHT CET 2020 · 16 OCT · SHIFT 2Easy

Electric Flux & Gauss's Law

A spherical rubber balloon carries a charge, uniformly distributed over the surface. As the balloon is blown up and increases in size, the total electric flux coming out of the surface

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Answer: (b)

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Q91JEE Main 2020 · 8 JAN · SHIFT 1Easy

Electric Flux & Gauss's Law

In finding the electric field using Gauss Law the formula ∣E⃗∣=qencε0∣A∣\left|\vec{E}\right| = \frac{q_{enc}}{\varepsilon_0 |A|} is applicable. In the formula ε0\varepsilon_0 is permittivity of free space, A is the area of Gaussian surface and qenc is charge enclosed by the Gaussian surface. The equation can be used in which of the following situation?

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Answer: (d)

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Q92JEE Main 2020 · 6 SEP · SHIFT 2Easy

Electric Flux & Gauss's Law

Consider the force F on a charge 'q' due to a uniformly charged spherical shell of radius R carrying charge Q distributed uniformly over it. Which one of the following statements is true for F, if 'q' is placed at distance r from the centre of the shell?

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Answer: (b)

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Q93MHT CET 2020 · 19 OCT · SHIFT 2Easy

Electric Flux & Gauss's Law

Van de Graaff generator produces

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Answer: (b)

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Q94NEET 2020 · PHASE 1Easy

Electric Flux & Gauss's Law

A spherical conductor of radius 10 cm has a charge of 3.2 ×\times 10-7 C distributed uniformly. What is the magnitude of electric field at a point 15 cm from the centre of the sphere?

(14πε0=9×109 N m2/C2)\left(\frac{1}{4\pi\varepsilon_0} = 9 \times 10^{9}\text{ N m}^2/\text{C}^2\right)

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Q95JEE Main 2020 · 9 JAN · SHIFT 2Medium

Electric Flux & Gauss's Law

An electric field E⃗=4xi^−(y2+1)j^\vec{E} = 4x\hat{i} - (y^{2} + 1)\hat{j} N/C

passes through the box shown in figure. The

flux of the electric field through surfaces ABCD

and BCGF are marked as ϕI\phi_{I} and ϕII\phi_{II}

respectively. The difference between (ϕI−ϕII)(\phi_{I} - \phi_{II}) is (in Nm2/C) _______.

Figure

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Answer: -48

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Q96JEE Main 2020 · 7 JAN · SHIFT 1Medium

Electric Flux & Gauss's Law

Two infinite planes each with uniform surface charge density σ\sigma are kept in such a way that the angle between them is 30o. The electric field in the region shown between them is given by:

Figure

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Answer: (d)

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Q97MHT CET 2020 · 16 OCT · SHIFT 1Medium

Electric Flux & Gauss's Law

Surface density of charge on a charged conducting sphere of radius RR in terms of electric field intensity EE at a distance rr in free space is (r>R,ε0=r > R, \varepsilon_0 = permittivity of free space)

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Answer: (c)

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Q98JEE Main 2020 · 9 JAN · SHIFT 1Difficult

Electric Flux & Gauss's Law

Consider a sphere of radius R which carries a uniform charge density ρ\rho. If a sphere of radius R2\frac{R}{2} is carved out of it, as shown, the ratio Figure of magnitude of electric field E⃗A\vec{E}_A and E⃗B\vec{E}_B, respectively, at points A and B due to the remaining portion is:

Figure

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Answer: (c)

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Q99NEET 2019Easy

Electric Flux & Gauss's Law

A sphere encloses an electric dipole with charges ±3×10−6 C\pm 3 \times 10^{-6}\text{ C}. What is the total electric flux across the sphere?

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Q100NEET 2019Easy

Electric Flux & Gauss's Law

A hollow metal sphere of radius R is uniformly charged. The electric field due to the sphere at a distance r from the centre:

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Answer: (a)

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Q101NEET 2019Medium

Electric Flux & Gauss's Law

Two parallel infinite line charges with linear charge densities +λ\lambda C/m and -λ\lambda C/m are placed at a distance of 2R in free space. What is the electric field mid-way between the two line charges?

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Q102JEE Main 2019 · 11 JAN · SHIFT 1Medium

Electric Flux & Gauss's Law

The given graph shows variation (with distance r from centre) of:

Figure

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Answer: (d)

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Q103MHT CET 2019 · 3 MAY · SHIFT 1Medium

Electric Flux & Gauss's Law

A charged conductor produces an electric field of intensity 103 V/m10^{3}\text{ V/m} just outside its surface in vacuum. Then, it produces the electric field of intensity E just outside its surface, when it is placed in a medium of dielectric constant 4. The value of EE will be

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Answer: (c)

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Q104JEE Main 2019 · 12 APR · SHIFT 1Medium

Electric Flux & Gauss's Law

Shown in the figure is a shell made of a conductor. It has inner radius a and outer radius b, and carries charge Q. At its centre is a dipole P⃗\vec{P}as shown. In this case:

Figure

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Answer: (d)

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Q105JEE Main 2019 · 12 APR · SHIFT 2Difficult

Electric Flux & Gauss's Law

Let a total charge 2Q be distributed in a sphere of radius R, with the charge density given by ρ\rho(r) = kr, where r is the distance from the centre. Two charges A and B, of –Q each, are placed on diametrically opposite points, at equal distance, aa from the centre. If A and B do not experience any force, then:

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Answer: (a)

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Q106JEE Main 2019 · 9 JAN · SHIFT 2Difficult

Electric Flux & Gauss's Law

Charge is distributed within a sphere of radius R with a volume charge density ρ(r)=Ar2e−2ra,\rho(r) = \frac{A}{r^{2}}e^{-\frac{2r}{a}}, where A and a are constants. If Q is the total charge of this charge distribution, the radius R is:

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Answer: (b)

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Q107JEE Main 2018 · 15 APR · SHIFT 1Medium

Electric Flux & Gauss's Law

A charge QQ is placed at a distance a/2a/2 above the center of the square surface of edge a as shown in the figure.

Figure

The electric flux through the square surface is

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Answer: (d)

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Q108JEE Main 2018 · 15 APR · SHIFT 2Difficult

Electric Flux & Gauss's Law

A solid ball of radius R has a charge density ρ\rho

given by ρ=ρ\rho = \rhoo (1 −rR-\frac{r}{R}) for 0 ≤\leq r ≤\leq R. The electric field outside the ball is:

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Answer: (b)

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Q109JEE Main 2017 · 9 APR · SHIFT 1Easy

Electric Flux & Gauss's Law

Four closed surfaces and corresponding charge distributions are shown below.

Figure

Let the respective electric fluxes through the surfaces be Φ1,Φ2,Φ3\Phi_1, \Phi_2, \Phi_3 and Φ4\Phi_4. Then:

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Answer: (c)

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Q110JEE Main 2016Difficult

Electric Flux & Gauss's Law

The region between two concentric spheres of radii aa and b,b, respectively (see figure), has volume charge density ρ=Ar,\rho = \frac{A}{r}, where AA is a constant and rr is the distance from the center. At the center of the spheres is a point charge QQ. The value of AA such that the electric field in the region between the spheres will be constant, is:

Figure

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Answer: (c)

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Q111NEET 2015 · AIPMT · CANCELLED PAPERMedium

Electric Flux & Gauss's Law

The electric field in a certain region is acting radially outward and is given by E = Ar. A charge contained in a sphere of radius 'a' centred at the origin of the field, will be given by

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Answer: (a)

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Q112JEE Main 2015 · 10 APR · SHIFT 1Medium

Electric Flux & Gauss's Law

Shown in the figure are two point charges + Q and −- Q inside the cavity of a spherical shell. The charges are kept near the surface of the cavity on opposite sides of the centre of the shell. If σ\sigma1 is the surface charge on the inner surface and Q1 net charge on it and σ\sigma2 the surface charge on the outer surface and Q2 net charge on it then:

Figure

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Answer: (c)

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Q113JEE Main 2014 · 19 APR · SHIFT 1Medium

Electric Flux & Gauss's Law

The electric field in a region of space is given by, E⃗=E0i^\vec{E} = E_0\hat{i} + 2E0j^E_0\hat{j} where E0 = 100 N/C. The flux of this field through a circular surface of a radius 0.02 m parallel to the Y-Z plane is nearly:

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Answer: (a)

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Q114JEE Main 2014 · 11 APR · SHIFT 1Medium

Electric Flux & Gauss's Law

A cone of base radius R and height h is located in a uniform electric field E⃗\vec{E}parallel to its base. The electric flux entering the cone is:

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Answer: (b)

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Q115JEE Main 2014 · 9 APR · SHIFT 1Medium

Electric Flux & Gauss's Law

The magnitude of the average electric field normally present in the atmosphere just above the surface of the Earth is about 150 N/C, directed inward towards the center of the Earth. This gives the total net surface charge carried by the Earth to be:

[Given ϵ\epsilon0 = 8.85 ×\times 10−-12 C2/N-m2, RE = 6.37 ×\times 106 m]

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Answer: (c)

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Q116JEE Main 2014 · 12 APR · SHIFT 1Difficult

Electric Flux & Gauss's Law

A spherically symmetric charge distribution is characterised by a charge density having the following variation:

ρ\rho(r) = ρ\rho0 (1−rR)\left(1 - \frac{r}{R}\right) for r < R

ρ\rho(r) = 0 for r ≥\geq R

Where r is the distance from the center of the charge distribution and ρ\rho0 is a constant. The electric field at an internal point (r < R) is:

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Answer: (b)

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Q117NEET 2012 · AIPMT · PRELIMSEasy

Electric Flux & Gauss's Law

What is the flux through a cube of side aa if a point charge of q is at one of its corners?

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Answer: (b)

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Q118JEE Main 2012 · AIEEEEasy

Electric Flux & Gauss's Law

In a uniformly charged sphere of total charge QQ and radius R,R, the electric field EE is plotted as a function of distance from the center. The graph which would correspond to the above will be:

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Answer: (c)

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Q119JEE Main 2012 · AIEEEMedium

Electric Flux & Gauss's Law

This question has statement- 11 and statement- 2.2. Of the four choices given after the statements, choose the one that best describes the two statements.

An insulating solid sphere of radius RR has a uniformly positive charge density ρ\rho. As a result of this uniform charge distribution there is a finite value of electric potential at the center of the sphere, at the surface of the sphere and also at a point outside the sphere. The electric potential at infinity is zero.

Statement- 1:1: When a charge qq is taken from the centre to the surface of the sphere its potential energy changes by qρ3ε0\frac{q\rho}{3\varepsilon_0}

Statement- 2:2: The electric field at a distance r (r<R)r\,(r < R) from the center of the sphere is ρr3ε0\frac{\rho r}{3\varepsilon_0}

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Answer: (c)

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Q120NEET 2011 · AIPMT · PRELIMSEasy

Electric Flux & Gauss's Law

A charge Q is enclosed by a Gaussian spherical surface of radius R. If the radius is doubled, then the outward electric flux will

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Answer: (c)

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Q121NEET 2010 · AIPMT · PRELIMSEasy

Electric Flux & Gauss's Law

A square surface of side L meter in the plane of the paper is placed in a uniform electric field EE(volt/m) acting along the same plane at an angle θ\theta with the horizontal side of the square as shown in figurre.

Figure

The electric flux linked to the surface, in units of volt m is

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Answer: (d)

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Q122NEET 2010 · AIPMT · MAINSEasy

Electric Flux & Gauss's Law

The electric field at a distance 3R2\frac{3R}{2} from the centre of a charged conducting spherical shell of radius R is E. The electric field at a distance R2\frac{R}{2} from the centre of the sphere is

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Answer: (a)

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Q123JEE Main 2010 · AIEEEDifficult

Electric Flux & Gauss's Law

Let there be a spherically symmetric charge distribution with charge density varying as ρ(r)=ρ0(54−rR)\rho(r) = \rho_0\left(\frac{5}{4} - \frac{r}{R}\right) upto r=R,r = R, and ρ(r)=0\rho(r) = 0 for r>R,r > R, where rr is the distance from the origin. The electric field at a distance r (r<R)r\ (r < R) from the origin is given by

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Answer: (a)

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Q124JEE Main 2009 · AIEEEMedium

Electric Flux & Gauss's Law

Let ρ(r)=QπR4r\rho(r) = \frac{Q}{\pi R^{4}}r be the charge density distribution for a solid sphere of radius RR and total charge QQ. For a point pp inside the sphere at distance r1r_1 from the center of the sphere, the magnitude of electric field is:

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Answer: (b)

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Q125JEE Main 2008 · AIEEEEasy

Electric Flux & Gauss's Law

A thin spherical shell of radius RR has charge QQ spread uniformly over its surface. Which of the following graphs most closely represents the electric field E(r)E(r) produced by the shell in the range 0≤r<∞,0 \le r < \infty, where rr is the distance from the center of the shell?

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Answer: (a)

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Q126NEET 2007 · AIPMTEasy

Electric Flux & Gauss's Law

A hollow cylinder has a charge q coulomb within it. If ϕ\phi is the electric flux in units of volt-metre associated with the curved surface B, the flux linked with the plane surface A in units of V-m will be

Figure

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Answer: (d)

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Q127NEET 2006 · AIPMTEasy

Electric Flux & Gauss's Law

A square surface of side L metres is in the plane of the paper. A uniform electric field E⃗\vec{E}(volt/m), also in the plane of the paper is limited only to the lower half of the square surface (see figure). The electric flux in SI inits associated with the surface is

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Answer: (d)

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Q128NEET 2003 · AIPMTEasy

Electric Flux & Gauss's Law

A charge q is located at the centre of a cube. The electric flux through any face is

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Answer: (b)

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Q129JEE Main 2003 · AIEEEEasy

Electric Flux & Gauss's Law

If the electric flux entering and leaving an enclosed surface respectively is ϕ1\phi_1 and ϕ2,\phi_2, the electric charge inside the surface will be

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Answer: (a)

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Q130JEE Main 2002 · AIEEEEasy

Electric Flux & Gauss's Law

A charged particle qq is placed at the centre OO of cube of length L(ABCDEFGH).L(ABCDEFGH). Another same charge qq is placed at a distance LL from OO. Then the electric flux through ABCDABCD is

Figure

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Answer: (b)

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Q131NEET 2001 · AIPMTEasy

Electric Flux & Gauss's Law

A charge Qμ\muC is placed at the centre of a cube, the flux coming out from each face will be

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Answer: (a)

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Q132NEET 2000 · AIPMTEasy

Electric Flux & Gauss's Law

A charge Q is situated at the corner of a cube, the electric flux passed through all the six faces of the cube is

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Answer: (b)

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Q133NEET 1998 · AIPMTEasy

Electric Flux & Gauss's Law

A hollow insulated conducting sphere is given a positive charge of 10 μC10\ \mu\text{C}. What will be the electric field at the centre of the sphere if its radius is 2 m2\text{ m}?

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Answer: (a)

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Q134NEET 1996 · AIPMTEasy

Electric Flux & Gauss's Law

A point charge +q+q is placed at the centre of a cube of side ll. The electric flux emerging from the cube is

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Answer: (d)

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