Question

The distance between electric charges QC and $9$QC is $4$m. What is the electric potential at a point of the line joining them where the electric field is zero?

• A. $4$ kQ v
• B. $10$ kQ v
• C. $2$ kQ v
• D. $2.5$ kQ v

Answer 1

The correct option is A $4$ kQ v

We know that Electric field and electric potential due to charge Q at a distance r is given by $\frac{KQ}{r^2}$ and $\frac{KQ}{r}$ respectively.

Let the point (say P) where electric field is zero is x distance apart from charge QC and (4-x) distance apart from charge 9QC as shown in figure.

since the direction of electric field is opposite, their magnitude must be equal in order to get zero electric field.

So, $\frac{KQ}{x^2}$ = $\frac{9KQ}{(4-x{)}^2}$

$\Rightarrow$ $\frac{1}{x^2}$ = $\frac{9}{(4-x{)}^2}$

$\Rightarrow$ $\frac{1}{x}$ = $\frac{3}{(4-x)}$$\Rightarrow$ $x=1m$.

Now, Electric potential at point P= Potential due charge QC + Potential due charge 9QC

$\Rightarrow$ $\frac{KQ}{1}$ + $\frac{9KQ}{3}$ = $4KQv$.

Therefore, A is correct option.