Question

Four charges are placed on the vertices of a square as shown in figure. The net electric potential at centre of the square is

• A. $10\text{V}$
• B. $15\text{V}$
• C. $29\text{V}$
• D. $\text{Zero}$

Answer 1

The correct option is D $\text{Zero}$

Let $a$ be the side length of the square.

So, distance between vertex and centre of square is $\frac{a}{\sqrt{2}}$

Now, we know potential due to a system of point charges at any point $\text{P}$ is given by

$V_p=\frac{k{q}_1}{r_1}+\frac{k{q}_2}{r_2}+\frac{k{q}_3}{r_3}+......+\frac{k{q}_n}{r_n}$

Since the all charges are kept at equal distance from centre point $\text{P}$ which is $(r=\frac{a}{\sqrt{2}})$, then

$\therefore V_p$ = $\frac{k}{r}(q_1+q_2+q_3+.....+q_n)$

$\Rightarrow V_p=\frac{k{q}_{total}}{r}$

Here, from centre of the square, all charges are equidistant.

$q_{total}=+4+(-4)+2+(-2)=0$

So, $V_p=0$

Hence, option $\left(a\right)$ is the correct answer.

Why this question ? Concept - If all charges are kept at equal distance $\left(r\right)$ from the point where $V$ is to be evaluated, then we can write, $V=\frac{k{q}_{total}}{r}$.