Question

# Charges of $$5 \times 10^{-6}$$ and $$-3 \times 10^{-6} C$$ are kept 16 cm apart. Find the position of points on the line joining the charges where the electric potential is zero.

Solution

## Potential at P iss zero$$\dfrac{k(5 \times 10^{-6})}{x} = \dfrac{k(3 \times 10^{-6})}{0.16-x}$$$$5(0.16 -x) = 3x$$$$0.8 = 8x$$$$x = 0.1m$$$$x = 10cm$$Physics

Suggest Corrections

0

Similar questions
View More

People also searched for
View More