A cubical region of side $a$ has its centre at the origin. It encloses three fixed point charges, $-q$ at $(0,-a / 4,0),+3 q$ at $(0,0,0)$ and $-q$ at $(0,+a / 4,0)$. Choose the correct option(s).

(A) The net electric flux crossing the plane $x=+a / 2$ is equal to the net electric flux crossing the plane $x=-a / 2$.; (C) The net electric flux crossing the entire region is $\frac{q}{\varepsilon_{0}}$.; (D) The net electric flux crossing the plane $z=+a / 2$ is equal to the net electric flux crossing the plane $x=+a / 2$.

(B) The net electric flux crossing the plane $y=+a / 2$ is more than the net electric flux crossing the plane $y=-a / 2$.

JEE Advanced · 2012 · 08 Apr 2012 · Paper 1 · Electrostatics

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A cubical region of side

a

has its centre at the origin. It encloses three fixed point charges,

- q

(0, - a /4, 0), + 3 q

(0, 0, 0)

and

- q

(0, + a /4, 0)

. Choose the correct option(s).

(A)The net electric flux crossing the plane $x = + a /2$ is equal to the net electric flux crossing the plane $x = - a /2$ .✓ Correct
(B)The net electric flux crossing the plane $y = + a /2$ is more than the net electric flux crossing the plane $y = - a /2$ .
(C)The net electric flux crossing the entire region is $\frac{q}{ε _{0}}$ .✓ Correct
(D)The net electric flux crossing the plane $z = + a /2$ is equal to the net electric flux crossing the plane $x = + a /2$ .✓ Correct

Answer: (A), (C), (D)

Solution

The net charge enclosed is

- q + 3 q - q = q

, so by Gauss's law the total flux is

\frac{q}{ε _{0}}

. Due to symmetry about the

x

and

z

axes, the flux through

x = + a /2

equals that through

x = - a /2

, and similarly for the

z

planes. The charge distribution is symmetric about

y = 0

, so the flux through

y = + a /2

equals that through

y = - a /2

, not more. Options (A), (C), and (D) are correct.

JEE Advanced · 2012 · 08 Apr 2012 · Paper 1 · One or more correct options

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JEE Advanced · 2012 · 08 Apr 2012 · Paper 1 · Electrostatics

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