Two charges `3 xx 10^(-8)` C and `- 2 xx 10^(-8)` C are located 15 cm apart. At what point on the line joining the two charges is the electric potential zero ? Take the potential at infinity to be zero.

Explanation

Suppose charge of `3 xx 10^(-8)` C be taken on the origin O and charge of `- 2 xx 10^(-8)` C be taken on point A along positive x-axis at distance 15 cm as shown in figure.

The potential at point P is zero. As seen logically the potential due to both charges cannot be zero for left side of O (x `lt` 0) and hence there are two possibilities for point P.
(i) If point P is between O and A, then suppose P is from O at distance x cm and hence point A is at distance (15 - x) cm from P. Potential at P due to charge on point O is
`V_(1) =(k(3xx10^(-8)))/(x xx 10^(-2))` and
Potential at P due to charge on point A is
`V_(2)=(k(-2xx10^(-8)))/((15-x)xx10^(-2))`

but potential at P is
`V=V_(1)+V_(2)`
`O=(kxx3xx10^(-8))/(x xx10^(-2))-(kxx2xx10^(-8))/((15-x)xx10^(-2))`
`O =(3)/(x)-(2)/(15-x)`
`:. (2)/(15-x)=(3)/(x)`
`:. 2x= 45 -3x`
`:. 5x=45`
`:. x=9` cm
(ii) If neutral point obtain from right side of A them .

`:.` Now potential at P due to charge on point O is
`V_(1)=(k(3xx10^(-8)))/(x xx10^(-2))`
Potential at P due to charge on point A
`V_(2) =(k(-2xx10^(-8)))/((x-15)xx10^(-2))=-(k(2xx10^(-8)))/((x-15)xx10^(-2))`
Total potential at point P,
`V= V_(1)+V_(2)`
`:. O = (k(3xx10^(-8)))/(x xx10^(-2))-(k(2xx10^(-8)))/((x-15)xx10^(-2))`
`:. (3)/(x)=(2)/(x-15)`
`:. 3x-45=2x`
`:. x= 45 cm`
Thus, electric potential is zero at 9 cm and 45 cm away from the positive charge on the side of the negative charge .
Note : Here, choosing a potential to be zero at infinity.