Question

A charged particle having a charge of $-2.o\times {10}^{-6}C$ is placed close to a nonconducting plate having a surface charge density $4.0\times {10}^{-6}C{m}^{-2}$. Find the force of attraction between the particle and the plate.

Answer 1

We know that E due to conducting charge density

$E=\frac{\sigma }{2{\in }_0}$

$ThenF=qE=\frac{q\sigma }{2{\in }_0}$

$=\frac{2.0\times {10}^{-6}\times 4.0\times {10}^{-6}}{2\times 8.8\times {10}^{-12}}$

$=0.45N.$