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Electrostatics question

2023 · 1 Feb · Shift 2 · Q74
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Electrostatics question

2023 · 1 Feb · Shift 2 · Q74

JEE MainPhysicsElectrostaticsNumerical+4 / −1
A cubical volume is bounded by the surfaces x=0,x=a,y=0,y=a,z=0,z=a\mathrm{x}=0, x=\mathrm{a}, y=0, y=\mathrm{a}, \mathrm{z}=0, z=\mathrm{a}x=0,x=a,y=0,y=a,z=0,z=a. The electric field in the region is given by E→=E0xi^\overrightarrow{\mathrm{E}}=\mathrm{E}_{0} x \hat{i}E=E0​xi^. Where E0=4×104 NC−1 m−1\mathrm{E}_{0}=4 \times 10^{4} ~\mathrm{NC}^{-1} \mathrm{~m}^{-1}E0​=4×104 NC−1 m−1. If a=2 cm\mathrm{a}=2 \mathrm{~cm}a=2 cm, the charge contained in the cubical volume is Q×10−14C\mathrm{Q} \times 10^{-14} \mathrm{C}Q×10−14C. The value of Q\mathrm{Q}Q is ‾\underline{\hspace{2cm}}​. (Take ϵ0=9×10−12 C2/Nm2\epsilon_{0}=9 \times 10^{-12} ~\mathrm{C}^{2} / \mathrm{Nm}^{2}ϵ0​=9×10−12 C2/Nm2)
Numerical answer
View written solutionFree

Correct answer: 288

  1. Use Gauss's law in differential/integral form

For a closed surface,

∮E⃗⋅dA⃗=qencε0\oint \vec E\cdot d\vec A=\frac{q_{\text{enc}}}{\varepsilon_0}∮E⋅dA=ε0​qenc​​

So,

qenc=ε0∮E⃗⋅dA⃗q_{\text{enc}}=\varepsilon_0\oint \vec E\cdot d\vec Aqenc​=ε0​∮E⋅dA

Given,

E⃗=E0x i^\vec E=E_0 x\,\hat iE=E0​xi^

This field is along the xxx-direction only, so only the faces perpendicular to the xxx-axis contribute to flux.

  1. Flux through each face

The cube is bounded by:

x=0,  x=a,  y=0,  y=a,  z=0,  z=ax=0,\; x=a,\; y=0,\; y=a,\; z=0,\; z=ax=0,x=a,y=0,y=a,z=0,z=a

with side length aaa.

Face at x=0x=0x=0

Here,

E⃗=E0(0)i^=0\vec E=E_0(0)\hat i=0E=E0​(0)i^=0

Hence flux through this face is

Φx=0=0\Phi_{x=0}=0Φx=0​=0

Face at x=ax=ax=a

Here,

E⃗=E0a i^\vec E=E_0 a\,\hat iE=E0​ai^

Outward area vector is along +i^+\hat i+i^, and area is a2a^2a2. So,

Φx=a=E0a⋅a2=E0a3\Phi_{x=a}=E_0 a\cdot a^2=E_0 a^3Φx=a​=E0​a⋅a2=E0​a3

Faces at y=0,y=a,z=0,z=ay=0,y=a,z=0,z=ay=0,y=a,z=0,z=a

Their area vectors are perpendicular to i^\hat ii^, so

E⃗⋅dA⃗=0\vec E\cdot d\vec A=0E⋅dA=0

Thus their flux is zero.

  1. Total flux

Therefore,

Φ=E0a3\Phi=E_0 a^3Φ=E0​a3

So enclosed charge is

qenc=ε0E0a3q_{\text{enc}}=\varepsilon_0 E_0 a^3qenc​=ε0​E0​a3
  1. Substitute values

Given:

E0=4×104  N C−1m−1E_0=4\times 10^4\; \text{N C}^{-1}\text{m}^{-1}E0​=4×104N C−1m−1 a=2 cm=2×10−2 ma=2\text{ cm}=2\times 10^{-2}\text{ m}a=2 cm=2×10−2 m ε0=9×10−12  C2/N m2\varepsilon_0=9\times 10^{-12}\; \text{C}^2\text{/N m}^2ε0​=9×10−12C2/N m2

Now,

a3=(2×10−2)3=8×10−6a^3=(2\times 10^{-2})^3=8\times 10^{-6}a3=(2×10−2)3=8×10−6

Then,

qenc=(9×10−12)(4×104)(8×10−6)q_{\text{enc}}=(9\times 10^{-12})(4\times 10^4)(8\times 10^{-6})qenc​=(9×10−12)(4×104)(8×10−6)

First multiply the numbers:

9×4×8=2889\times 4\times 8=2889×4×8=288

And powers of 10:

10−12×104×10−6=10−1410^{-12} \times 10^4 \times 10^{-6}=10^{-14}10−12×104×10−6=10−14

Hence,

qenc=288×10−14 Cq_{\text{enc}}=288\times 10^{-14}\text{ C}qenc​=288×10−14 C

So in the form Q×10−14Q\times 10^{-14}Q×10−14 C,

Q=288Q=288Q=288
  1. Comparison with stored answer

Stored correct answer: 288288288

Our derived answer is also 288288288, so it agrees.

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