Four identical positive point charge q are placed at the corner of a square plate of side ‘d’. Another point fixed charge q is placed at a distance ‘d/2’ from the centre of square as shown. Choose the correct options.

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Electrostatic potential energy of the system is $\frac{\sqrt{2} k q^{2}}{d} [1 + 2 \sqrt{2} + 4 \sqrt{\frac{2}{3}}]$

Flux through the square plate is $\frac{q}{2 ε_{0}}$

Flux through the square plate is $\frac{q}{6 ε_{0}}$

Work done in moving a charge $q_{0}$ from infinity to the centre of the square plate with charges q each at the vertices is $2 (2 \sqrt{2} + 1) \frac{k q q_{0}}{d}$

answer is A.

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Detailed Solution

$\begin{array}{l} U_{s y s} = \frac{4 k q^{2}}{d} + \frac{2 k q^{2}}{\sqrt{2} d} + \frac{8 k q^{2}}{\sqrt{3} d} = \frac{\sqrt{2} k q^{2}}{d} [2 \sqrt{2} + 1 + \frac{4 \sqrt{2}}{\sqrt{3}}] \\ P o t e n t i a l a t t h e c e n t r e \\ V_{f} = \frac{4 \sqrt{2} k q}{d} + \frac{2 k q}{d} \\ W o r k d o n e = q_{0} [V_{f} - V_{i}] = 2 (2 \sqrt{2} + 1) \frac{k q q_{0}}{d} \\ T h e f l u x o f c h a r g e a t c o r n e r s o f s q u a r e i s z e r o . \\ I m a g i n e a c u b e o f s i d e d, w i t h q a t i t s c e n t e r . \\ ϕ_{c u b e} = \frac{q}{ε_{0}} \\ \Rightarrow ϕ_{f a c e} = \frac{q}{6 ε_{0}} \end{array}$