Question

If E be the electric field inside a PPC due to Q and -Q charges on the two plates, then electrostatic force on plates If E be the electric field inside a PPC due to Q and -Q charges on the two plates, then electrostatic force on plates

Answer 1

Answer:

See below.

Explanation:

We know that the electric field present in the parallel plate capacitor is,

$E=\dfrac {\sigma }{2\varepsilon _{0}}$

Hence, the force on a charge of Q will be,

$\begin{aligned}F=QE\\ =\dfrac {Q\sigma }{2\varepsilon _{0}}\end{aligned}$

Which can also be written as,

$\begin{aligned}F=\dfrac {Q\left( \dfrac {Q}{A}\right) }{2\varepsilon _{0}}\\ =\dfrac {Q^{2}}{2A\varepsilon _{0}}\end{aligned}$