Question
Identical charges $$\left( { - q} \right)$$ are placed at each corners of a cube of side $$b,$$ then the electrostatic potential energy of charge $$\left( { + q} \right)$$ placed at the centre of the cube will be
A.
$$ - \frac{{4\sqrt 2 {q^2}}}{{\pi {\varepsilon _0}}}$$
B.
$$\frac{{8\sqrt 2 {q^2}}}{{\pi {\varepsilon _0}b}}$$
C.
$$ - \frac{{4{q^2}}}{{\sqrt 3 \pi {\varepsilon _0}b}}$$
D.
$$\frac{{8\sqrt 2 {q^2}}}{{4\pi {\varepsilon _0}b}}$$
Answer :
$$ - \frac{{4{q^2}}}{{\sqrt 3 \pi {\varepsilon _0}b}}$$
Solution :
Electrostatic potential energy of charge $$+q$$ placed at the centre of cube is $$U = 8 \times \frac{1}{{4\pi {\varepsilon _0}}} \times \frac{{q\left( { - q} \right)}}{{{\text{half - diagonal distance}}}}$$
$$\eqalign{
& = 8 \times \frac{1}{{4\pi {\varepsilon _0}}}\frac{{ - {q^2}}}{{b\frac{{\sqrt 3 }}{2}}}\left[ {_{{\text{where,}}\,b\, = \,{\text{side}}\,{\text{of}}\,{\text{cube}}}^{{\text{diagonal}}\,{\text{of}}\,{\text{cube}}\,\,{\text{ = }}\sqrt 3 b}} \right] \cr
& = \frac{{ - 4{q^2}}}{{\sqrt 3 \pi {\varepsilon _0}b}} \cr} $$