Question 569: Electrostatics & Capacitance

Question

In parallel plate capacitor, electric field between the plates is '$E$'. If the charge on the plates is '$Q$' then the force on each plate is

Accepted Answer

The field produced by charge on each plate is $\frac{E}{2}$.
Hence force on each plate is given by
$\mathrm{F}=\mathrm{Q} 	imes$ (Field produced by the other plate)
$=\frac{\mathrm{QE}}{2}$

Answer

QE

Answer

$\frac{\mathrm{QE}^2}{2}$

Answer

$\mathrm{QE}^2$

Answer

$\frac{\mathrm{QE}}{2}$

Electrostatics & Capacitance

In parallel plate capacitor, electric field between the plates is 'E'. If the charge on the plates is 'Q' then the force on each plate is

Choose the correct answer: