Question

Two identical positive charges aap placed on the y-axis at y is equal to minus a and y is equal to plus a. The variation of electric potential along x axis is shown by the graph

Answer 1

Answer:

(1)

Explanation:

Let there be positive charge at the positions (0,a) and (0,-a) and there exists a point P at a distance 'x' from origin on x-axis.

We need to find Electric Potential at point P,

In $\triangle AOP,$

Using Pythagoras Theorem,

$AO^2+OP^2=AP^2\\ \\a^2+x^2=OP^2\\ \\OP=\sqrt{a^2+x^2}$

Now, I need to find Electric Potential at that point due to two charges,

That is,

$V_{net}=\frac{Kq}{OP}+\frac{Kq}{OP}\\ \\= \frac{Kq}{\sqrt{a^2+x^2}}+\frac{Kq}{\sqrt{a^2+x^2}}\\ \\=\frac{2Kq}{\sqrt{a^2+x^2}}$

Hence, Our Expression of net Potential matches with option (1).

Verification:

$\lim_{x \to{ \pm \infty}} \frac{2Kq}{\sqrt{a^2+x^2}}=0\\ \\ \lim_{x \to 0} \frac{2Kq}{\sqrt{a^2+x^2}} =\frac{2Kq}{a}$