A cubical volume is bounded by the surfaces x = 0, x = a, y = 0, y = a, z = 0, z = a. The electric field in the region is given by E = E₀ xi. Where E₀ = 4 × 10⁴ NC⁻¹ m⁻¹. If a = 2 cm, the charge contained in the cubical volume is Q × 10⁻¹⁴ C. The value of Q is ________________. (Take ε₀ = 9 × 10⁻¹² C²/Nm²) Numerical answer type. Work it out, then open the answer.

Answer: 288. Only the faces x = 0 and x = a give flux; at x = 0, E = 0. φ = E₀a· a² = E₀a³ = 4×10⁴×(0.02)³ = 0.32 N m ² /C Q = ε₀φ = 9×10⁻¹²×0.32 = 2.88×10⁻¹² = 288×10⁻¹⁴ C So Q = 288.

A cubical volume is bounded by the surfaces x = 0, x = a, y = 0, y =… | JEE Main 2023 (1 Feb, Shift 2)

Q65JEE Main 2023 · 1 FEB · SHIFT 2Medium

Electric Flux & Gauss's Law

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

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

Show answer and solution

Answer: 288

Only the faces $x = 0$ and $x = a$ give flux; at $x = 0$ , $E = 0$ .
$\phi = E_0a\cdot a^2 = E_0a^3 = 4\times10^{4}\times(0.02)^3 = 0.32$ N m $^2$ /C
$Q = \varepsilon_0\phi = 9\times10^{-12}\times0.32 = 2.88\times10^{-12} = 288\times10^{-14}$ C
So $Q = 288$ .

Report a mistake

Electric Flux & Gauss's Law question from JEE Main 2023 (1 Feb, Shift 2)

More questions on Electric Flux & Gauss's Law