Question 594: Electrostats & 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

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

Electrostats & 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

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