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

Explanation

Given, `q_(A)= 3 xx 10^(-8)C`,
`q_(B)= -2 xx 10^(-8)C and r= 15cm= 0.15m`
Let O be the point, where the electric potential is zero due to the two charges.

Suppose that the distance AO= x
Then, `BO= r- x= 0.15-x`
Electric potential at point O due to `q_(A)`
`V_(A)= (1)/(4pi epsi_(0)).(q_(A))/(AO) =9 xx 10^(9) xx (3 xx 10^(-8))/(x)= (270)/(x)`
Electric potential at point O due to `q_(B)`,
`V_(B)= (1)/(4pi epsi_(0)) (q_(B))/(BO) =9 xx 10^(9) xx ((-2 xx 10^(-8)))/(0.15-x)= -(180)/(0.15-x)`
Since the electric potential at point O is zero, we have `V_(A) + V_(B)= 0`
or `(270)/(x) + (-(180)/(0.15-x))= 0 or (270)/(x)= (180)/(0.15-x)`
or x= 0.09m =9cm (from charge `3 xx 10^(-8)C`)