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Electrostatics question

2023 · 30 Jan · Shift 2 · Q69
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Electrostatics question

2023 · 30 Jan · Shift 2 · Q69

JEE MainPhysicsElectrostaticsNumerical+4 / −1
As shown in figure, a cuboid lies in a region with electric field E=2x2i^−4yj^+6k^ N/CE=2 x^{2} \hat{i}-4 y \hat{j}+6 \hat{k} \mathrm{~N} / \mathrm{C}E=2x2i^−4yj^​+6k^ N/C. The magnitude of charge within the cuboid is n∈0Cn \in_{0} Cn∈0​C. The value of nnn is ‾\underline{\hspace{2cm}}​ (if dimension of cuboid is 1×2×3 m31 \times 2 \times 3 \mathrm{~m}^{3}1×2×3 m3 ) JEE Main 2023 (Online) 30th January Evening Shift Physics - Electrostatics Question 97 English
Numerical answer
View written solutionFree

Correct answer: 12

  1. Use Gauss’s law in differential form

For a volume charge distribution,

∇⋅E⃗=ρε0\nabla \cdot \vec E = \frac{\rho}{\varepsilon_0}∇⋅E=ε0​ρ​

So total charge inside the cuboid is

Q=∭ρ dV=ε0∭(∇⋅E⃗) dVQ=\iiint \rho\, dV=\varepsilon_0 \iiint (\nabla\cdot \vec E)\, dVQ=∭ρdV=ε0​∭(∇⋅E)dV
  1. Given electric field
E⃗=2x2 i^−4y j^+6 k^\vec E = 2x^2\,\hat i -4y\,\hat j + 6\,\hat kE=2x2i^−4yj^​+6k^

Now compute divergence:

∇⋅E⃗=∂∂x(2x2)+∂∂y(−4y)+∂∂z(6)\nabla\cdot \vec E = \frac{\partial}{\partial x}(2x^2)+\frac{\partial}{\partial y}(-4y)+\frac{\partial}{\partial z}(6)∇⋅E=∂x∂​(2x2)+∂y∂​(−4y)+∂z∂​(6) ∇⋅E⃗=4x−4+0=4x−4\nabla\cdot \vec E = 4x-4+0=4x-4∇⋅E=4x−4+0=4x−4
  1. Dimensions of cuboid

From the figure/data, take the cuboid spans:

0≤x≤1,0≤y≤2,0≤z≤30\le x\le 1,\qquad 0\le y\le 2,\qquad 0\le z\le 30≤x≤1,0≤y≤2,0≤z≤3

Hence,

Q=ε0∭(4x−4) dVQ=\varepsilon_0\iiint (4x-4)\,dVQ=ε0​∭(4x−4)dV
  1. Integrate over the cuboid
Q=ε0∫01∫02∫03(4x−4) dz dy dxQ=\varepsilon_0 \int_0^1 \int_0^2 \int_0^3 (4x-4)\,dz\,dy\,dxQ=ε0​∫01​∫02​∫03​(4x−4)dzdydx

First integrate over zzz and yyy:

∫03dz=3,∫02dy=2\int_0^3 dz = 3, \qquad \int_0^2 dy = 2∫03​dz=3,∫02​dy=2

So,

Q=ε0⋅6∫01(4x−4) dxQ=\varepsilon_0 \cdot 6 \int_0^1 (4x-4)\,dxQ=ε0​⋅6∫01​(4x−4)dx Q=6ε0[2x2−4x]01Q=6\varepsilon_0 \left[2x^2-4x\right]_0^1Q=6ε0​[2x2−4x]01​ Q=6ε0(2−4)=6ε0(−2)=−12ε0Q=6\varepsilon_0(2-4)=6\varepsilon_0(-2)=-12\varepsilon_0Q=6ε0​(2−4)=6ε0​(−2)=−12ε0​
  1. Magnitude of charge
∣Q∣=12ε0|Q|=12\varepsilon_0∣Q∣=12ε0​

Given ∣Q∣=nε0 C|Q| = n\varepsilon_0\,\text{C}∣Q∣=nε0​C,

n=12n=12n=12
  1. Final answer
12\boxed{12}12​

The derived answer matches the stored correct answer.

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