The electric field intensity on the surface of a solid charged sphere of radius ' r ' and volume charge density ' ρ ' is ( ε₀ = permittivity of free space) (a) (ρ r)/(3ε₀) (b) (ρ)/(4πε₀ r) (c) zero (d) (5ρ r)/(6ε₀)

Answer: (a). Gauss's law: E · 4π r² = (ρ · 4/3π r³)/(ε₀) E = (ρ r)/(3ε₀) So the answer is (a).

The electric field intensity on the surface of a solid charged sphere… | MHT CET 2023 (14 May, Shift 1)

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 ' $r$ ' and volume charge density ' $\rho$ ' is ( $\varepsilon_0 =$ permittivity of free space)

Show answer and solution

Answer: (a)

Gauss's law: $E \cdot 4\pi r^2 = \dfrac{\rho \cdot \frac{4}{3}\pi r^3}{\varepsilon_0}$
$E = \dfrac{\rho r}{3\varepsilon_0}$
So the answer is (a).

Report a mistake

More questions on Electric Flux & Gauss's Law

Graphical variation of electric field due to a uniformly charged insulating solid sphere…JEE Main 2023 (8 Apr, Shift 1)
Let σ be the uniform surface charge density of two infinite thin plane sheets shown in…JEE Main 2023 (1 Feb, Shift 1)
Which of the following statements is 'WRONG' for the conductors?MHT CET 2023 (13 May, Shift 1)
As shown in the figure, a point charge Q is placed at the centre of conducting spherical…JEE Main 2023 (30 Jan, Shift 2)
A cubical volume is bounded by the surfaces x = 0, x = a, y = 0, y = a, z = 0, z = a. The…JEE Main 2023 (1 Feb, Shift 2)
Expression for an electric field is given by E = 4000x²i V/m. The electric flux through…JEE Main 2023 (31 Jan, Shift 1)
In a cuboid of dimension 2L × 2L × L, a charge q is placed at the center of the surface '…JEE Main 2023 (29 Jan, Shift 1)
As shown in figure, a cuboid lies in a region with electric field E = 2x²i - 4yj + 6k…JEE Main 2023 (30 Jan, Shift 2)

Electric Flux & Gauss's Law question from MHT CET 2023 (14 May, Shift 1)

More questions on Electric Flux & Gauss's Law