# 2.1)  Two charges$5 \times 10^{-8}C$  and $-3 \times 10^{-8}C$ are located 16 cm apart. At what point(s) on the line joining the two charges is the electric potential zero? Take the potential at infinity to be zero.

P Pankaj Sanodiya

Given, two charge particles

$q_1= 5*10^{-8}C$

$q_2= -3*10^{-8}C$

The separation between two charged particle $d=16cm=0.16m$

Now, let's assume the point P between two charged particles where the electric potential is zero is x meter away from $q_1$ and $( 9-x)$ meter away from  $q_2$

So,

The potential at point P :

$V_p=\frac{kq_1}{x}+\frac{kq_2}{0.16-x}=0$

$V_p=\frac{k5*10^{-8}}{x}+\frac{k(-3*10^{-8})}{0.16-x}=0$

$\frac{k5*10^{-8}}{x}=-\frac{k(-3*10^{-8})}{0.16-x}$

$5(0.16-x)=3x$

$x=0.1m=10cm$

Hence the point between two charged particles where the electric potential is zero lies 10cm away from $q_1$ and 6 cm away from $q_2$

Now, Let's assume a point  Q which is outside the line segment joining two charges and having zero electric potential .let the point Q lie r meter away from $q_2$ and  (0.16+r) meter away from $q_1$

So electric potential at point Q = 0

$\frac{kq_1}{0.16+r}+\frac{kq_2}{r}=0$

$\frac{k5*10^{-8}}{0.16+r}+\frac{k(-3*10^{-8})}{r}=0$

$5r=3(0.16+r)$

$r=0.24m=24cm$

Hence the Second point where the electric potential is zero is 24cm away from $q_2$ and 40cm away from $q_1$

Exams
Articles
Questions