Two charges `3xx10^(-8) C and -2xx10^(-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

Let us take the origin O at the location of the positive charge. The line joining the two charges is taken to be the x-axis the negative charge is taken to be on the right side of the origin

Let P be the required point on the x -axis where the potential is zero . If x is the x-coordinate of P obviously x mus be positive . (There is no possibility of potentials due to the two charges adding up to zero for `x lt0` ). If x lies between O and A we have
`(1)/(4piepsilon_(0))[(3xx10^(-8))/(x xx10^(-2))-(2xx10^(-8))/((15-x)xx10^(-2))]=0`
where x is in cm .i.e
`(3)/(x) - (2)/(15-x)=0`
which gives x =9 cm.
If x lies on the extended line OA the required condition is
`(3)/(x)-(2)/(x-15) =0`
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 that the formula for potential used in the calculation required choosing potential to be zero at infinity.