Question

Consider the arrangement of charges on the vertices of a square as shown in the figure. The net electrostatic potential at the centre of the square will be

Answer 1

Let Net Electrostatic Potential be V, so net potential is addition of potential of individual charges.

$V = \dfrac{kq}{d} + \dfrac{kq}{d} + \dfrac{kq}{d} + \dfrac{kq}{d}$

• Here 'd' is distance of charge from centre.
• d = s/√2 (using Pythagoras' Theorem)

$\implies V =4 \times \dfrac{kq}{d}$

$\implies V =4 \times \dfrac{kq}{ \bigg( \dfrac{s}{ \sqrt{2} } \bigg)}$

$\implies V = \dfrac{4 \sqrt{2} kq}{s}$

So, final answer is:

$\boxed{ \sf V = \dfrac{4 \sqrt{2} kq}{s}}$

Answer 2

Explanation:

Let Net Electrostatic Potential be V, so net potential is addition of potential of individual charges.

V = \dfrac{kq}{d} + \dfrac{kq}{d} + \dfrac{kq}{d} + \dfrac{kq}{d}V=

d

kq

+

d

kq

+

d

kq

+

d

kq

Here 'd' is distance of charge from centre.

d = s/√2 (using Pythagoras' Theorem)

\implies V =4 \times \dfrac{kq}{d} ⟹V=4×

d

kq

\implies V =4 \times \dfrac{kq}{ \bigg( \dfrac{s}{ \sqrt{2} } \bigg)} ⟹V=4×

(

2

s

)

kq

\implies V = \dfrac{4 \sqrt{2} kq}{s}⟹V=

s

4

2

kq

So, final answer is:

\boxed{ \sf V = \dfrac{4 \sqrt{2} kq}{s}}

V=

s

4

2

kq