Logo
Jeeify
NVAEasyJEE 2023Gauss's Law Applications

JEE Physics 2023 Question with Solution

A cubical volume is bounded by the surfaces x=0x = 0, x=ax = a, y=0y = 0, y=ay = a, z=0z = 0, z=az = a. The electric field in the region is given by E⃗=E0xi^\vec{E} = E_0 x \hat{i}. Where E0=4×104  N C−1m−1E_0 = 4 \times 10^4 \; \text{N C}^{-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/N m2\epsilon_0 = 9 \times 10^{-12} \, \text{C}^{2}/\text{N m}^{2})

Answer

Correct answer:288

Step-by-step solution

Standard Method

Given: E⃗=E0xi^\vec{E} = E_0 x \hat{i}, E0=4×104  N C−1m−1E_0 = 4 \times 10^4 \; \text{N C}^{-1} \text{m}^{-1}, a=2×10−2 ma = 2 \times 10^{-2} \, \text{m}, and ϵ0=9×10−12 C2/N m2\epsilon_0 = 9 \times 10^{-12} \, \text{C}^{2}/\text{N m}^{2}.

Find: The value of QQ if enclosed charge is written as Q×10−14 CQ \times 10^{-14} \, \text{C}.

The electric field is along the xx-direction, so only the two faces perpendicular to the xx-axis contribute to net flux. At x=0x = 0, the field is zero. At x=ax = a, the field is E⃗=E0ai^\vec{E} = E_0 a \hat{i}.

Using electric flux,

Φ=∫E⃗⋅dA⃗\Phi = \int \vec{E} \cdot d\vec{A}

so the net flux through the cube is

Φnet=E0a⋅a2=E0a3\Phi_{\text{net}} = E_0 a \cdot a^2 = E_0 a^3

Now apply Gauss's law,

Φnet=qenϵ0\Phi_{\text{net}} = \frac{q_{\text{en}}}{\epsilon_0}

Therefore,

qen=ϵ0E0a3q_{\text{en}} = \epsilon_0 E_0 a^3

Substituting the given values,

qen=(9×10−12)(4×104)(2×10−2)3q_{\text{en}} = \left(9 \times 10^{-12}\right) \left(4 \times 10^4\right) \left(2 \times 10^{-2}\right)^3 qen=36×10−8×8×10−6q_{\text{en}} = 36 \times 10^{-8} \times 8 \times 10^{-6} qen=288×10−14 Cq_{\text{en}} = 288 \times 10^{-14} \, \text{C}

Hence, in the form Q×10−14 CQ \times 10^{-14} \, \text{C}, the value of QQ is 288288.

Flux Through Relevant Faces

Given: E⃗=E0xi^\vec{E} = E_0 x \hat{i}.

Find: The enclosed charge inside the cube.

For the faces at y=0,y=a,z=0,z=ay = 0, y = a, z = 0, z = a, the area vectors are along ±j^\pm \hat{j} or ±k^\pm \hat{k}, while the field is along i^\hat{i}. Hence E⃗⋅A⃗=0\vec{E} \cdot \vec{A} = 0 on these four faces.

For the face at x=0x = 0,

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

so flux through this face is zero.

For the face at x=ax = a,

E⃗=E0ai^\vec{E} = E_0 a \hat{i}

and the outward area vector has magnitude a2a^2 in the +i^+\hat{i} direction. Therefore,

Φ=E⃗⋅A⃗=E0a⋅a2=E0a3\Phi = \vec{E} \cdot \vec{A} = E_0 a \cdot a^2 = E_0 a^3
Cubical volume with coordinate axes shown, shaded face at x equals a, and points A, B, C, D marked on that face.

Using Gauss's law,

qen=ϵ0Φnet=ϵ0E0a3q_{\text{en}} = \epsilon_0 \Phi_{\text{net}} = \epsilon_0 E_0 a^3

This gives

qen=288×10−14 Cq_{\text{en}} = 288 \times 10^{-14} \, \text{C}

Therefore, the required value is 288288.

Common mistakes

  • Considering flux through all six faces as nonzero. This is wrong because E⃗\vec{E} is along i^\hat{i}, so faces normal to j^\hat{j} and k^\hat{k} have zero dot product. Only faces perpendicular to the xx-axis need attention.

  • Forgetting that the field at x=0x = 0 is zero. This is wrong because E⃗=E0xi^\vec{E} = E_0 x \hat{i} depends on xx. Substitute x=0x = 0 first before computing flux through that face.

  • Using a=2a = 2 instead of converting 2 cm2 \, \text{cm} into 2×10−2 m2 \times 10^{-2} \, \text{m}. This gives the wrong power of ten. Always convert length to SI units before substitution.

Practice more Gauss's Law Applications questions

Get unlimited AI-adaptive practice, mastery tracking, and an AI tutor that explains every step — free to start.

Start free trialMore on Gauss's Law Applications

Related questions