Electrostatics

JEE Main level, single correct, medium

268 of 311

Question 268 of 311

Two point charges 2q2 q and qq are placed at vertex AA and centre of face CDEFC D E F of the cube as shown in figure. The electric flux passing through the cube is :

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Answer: (B) 3q4ϵ0\frac{3 q}{4 \epsilon_{0}}

Using Gauss's law:
ϕ=Qinϵ0\phi = \frac{Q_{in}}{\epsilon_0}

Charge 2q2q is at vertex A of the cube. A vertex is shared by 8 cubes, so the flux contribution from 2q2q through this cube is 2q8ϵ0=q4ϵ0\frac{2q}{8\epsilon_0} = \frac{q}{4\epsilon_0}.

Charge qq is at the centre of face CDEF. A face center is shared by 2 cubes, so the flux contribution from qq through this cube is q2ϵ0\frac{q}{2\epsilon_0}.

Total flux:
ϕ=q/4+q/2ϵ0=3q4ϵ0\phi = \frac{q/4 + q/2}{\epsilon_0} = \frac{3q}{4\epsilon_0}

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