In parallel plate capacitor, electric field between the plates is ' $\mathrm{E}$ '. If the charge on the…
In parallel plate capacitor, electric field between the plates is ' $\mathrm{E}$ '. If the charge on the plates is ' $Q$ ' then the force on each plate is
$\mathrm{QE}$
$\frac{\mathrm{QE}^2}{2}$
$\mathrm{QE}^2$
$\frac{\mathrm{QE}}{2}$
Solution
The field produced by charge on each plate is $\frac{E}{2}$.
Hence force on each plate is given by
$\mathrm{F}=\mathrm{Q} \times($ Field produced by the other plate $)$
$=\frac{\mathrm{QE}}{2}$