Exams › NEET › Physics

A parallel-plate air capacitor has capacitance C, plate separation d, and potential difference V is applied between the plates. The force of attraction between the plates is

1. \(\frac{CV^2}{2d}\)
2. \(\frac{CV}{2d}\)
3. \(\frac{CV^2}{d}\)
4. \(\frac{C^2V^2}{2d}\)

Correct answer: \(\frac{CV^2}{2d}\)

Solution

For a capacitor maintained at constant voltage, the attractive force can be obtained from the rate of change of energy with separation. Using \(C = \varepsilon_0 A/d\), the result is \(F = \frac{CV^2}{2d}\).

Related NEET Physics questions

• A capacitor is fully charged by a battery, and then the battery is disconnected. When a dielectric slab is inserted between the capacitor plates, what is the effect?
• In a charged capacitor, where is the energy stored?
• A capacitor of capacitance $C$ is charged to a potential $V$. The energy stored in it is:
• A capacitor of capacitance $50\,\mu\text{F}$ is charged to $10\,\text{V}$. What is the energy stored in it?
• A parallel-plate capacitor has air between its plates. When the air is replaced by a dielectric material of dielectric constant $K$, the capacitance:
• Eight identical small drops, each of radius $r$ and carrying charge $q$, coalesce to form one large drop. What is the ratio of the electric potential of the big drop to that of one small drop?

⚔️ Practice NEET Physics free + battle 1v1 →