JEE Main · 2021 · 24 Feb 2021 · Shift 1 · Electrostatics

A cube of side '

a

' has point charges

+ Q

located at each of its vertices except at the origin where the charge is -Q. The electric field at the centre of cube is:

(A) $\frac{2 Q}{3 3 π ε _{0} a ^{2}} (\overset{x}{^} + \overset{y}{^} + \overset{z}{^})$
(B) $\frac{Q}{3 3 π ε _{0} a ^{2}} (\overset{x}{^} + \overset{y}{^} + \overset{z}{^})$
(C) $\frac{- 2 Q}{3 3 π ε _{0} a ^{2}} (\overset{x}{^} + \overset{y}{^} + \overset{z}{^})$ ✓ Correct
(D) $\frac{- Q}{3 3 π ε _{0} a ^{2}} (\overset{x}{^} + \overset{y}{^} + \overset{z}{^})$

Answer: (C)

Solution

If all eight vertices had charge

+ Q

, the electric field at the center would be zero by symmetry.

The actual configuration is equivalent to having charge

- 2 Q

at the origin (since we have

- Q

instead of

+ Q

).

The distance from origin to center is

\frac{3}{2} a

. The electric field is:

E = \frac{1}{4 π ε _{0}} \frac{( - 2 Q )}{( \frac{3}{2} a ) ^{2}} \cdot \frac{1}{3} (\overset{x}{^} + \overset{y}{^} + \overset{z}{^}) = \frac{- 2 Q}{3 3 π ε _{0} a ^{2}} (\overset{x}{^} + \overset{y}{^} + \overset{z}{^})

JEE Main · 2021 · 24 Feb 2021 · Shift 1 · Single correct option