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MHT CET 202625 AprilEvening Shift
Exam
MHT CET
Year
2026
Date
25 April
Shift
Evening Shift
Subject
Physics
Chapter
Electrostatics

MHT CET 2026 · 25 April · Evening Shift · Physics · Electrostatics

A uniform electric field is given as E = 4 × 10^4\, j \ N/C . The...

A uniform electric field is given as E⃗=4×104 j^ N/C\vec{E} = 4 \times 10^4\,\hat{j}\ \text{N/C}. The flux of this field through a square of side 10 cm10\ \text{cm} whose plane is parallel to the Y-Z plane is

Options

  1. A400 Vm400\ Vm
  2. B200 Vm200\ Vm
  3. Czero
  4. D1600 Vm1600\ Vm

Correct answer

C. zero

Step-by-step solution

The electric field is given as E⃗=4×104 j^ N/C\vec{E} = 4 \times 10^4\,\hat{j}\ \text{N/C}.

The area of the square is A=(10 cm)2=0.01 m2A = (10\ \text{cm})^2 = 0.01\ \text{m}^2.

Since the plane of the square is parallel to the Y-Z plane, its area vector is perpendicular to the Y-Z plane, which means it is directed along the X-axis.

A⃗=0.01 i^ m2\vec{A} = 0.01\,\hat{i}\ \text{m}^2

The electric flux Φ\Phi through the square is given by the dot product of the electric field and the area vector:

Φ=E⃗⋅A⃗\Phi = \vec{E} \cdot \vec{A}

Φ=(4×104 j^)⋅(0.01 i^)\Phi = (4 \times 10^4\,\hat{j}) \cdot (0.01\,\hat{i})

Since j^⋅i^=0\hat{j} \cdot \hat{i} = 0, we get:

Φ=0\Phi = 0

Answer: zero
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