Question

Two charges $3\times {10}^{-8}$ and $2\times {10}^{-8}$ are located 15 cm apart.At what point on the line joining the two charges is the electric potential zero? Take potential at infinity to be zero.

Answer 1

Step 1, Given data and related figure

Two charges are

$Q_1=3\times {10}^{-8}$ C

$Q_2=2\times {10}^{-8}$ C

Distance between the charges = 15 cm

Step 2, Finding the point where the potential is zero

According to the figure let the potential is zero at point P

So,

$V_p=\frac{k{Q}_1}{x}+\frac{k{Q}_2}{15-x}=0$

Or,

$\frac{k{Q}_1}{x}=\frac{-k{Q}_2}{15-x}$

$Or,\frac{3\times {10}^{-8}}{x}=\frac{2\times {10}^{-8}}{15-x}$

After solving

x = 9 cm

Hence the potential is zero at 9 cm from $Q_1$ and 6 cm from $Q_2$