Name: Capacitance of a Parallel Plate Capacitor
Uploaded: 2019-07-19
Duration: 4 min 13 s
Description: Capacitors are available in various shapes and sizes. In this video, we will learn how to calculate the capacitance of a parallel plate capacitor.

Question

Derive an expression for the energy stored in a charged parallel-plate air capacitor. How does the energy stored change when the medium of air between the plates is replaced by a medium of dielectric constant

K

Explain.?

Accepted Answer

When a capacitor is charged, say by a battery, work is done by the charging battery (at the cost of its chemical energy). As the capacitor charges, the potential difference across its plates rises. More and more work has to be done by the battery in delivering the same amount of charge to the capacitor due to the rising potential difference across its plates. The total amount of work in charging the capacitor is stored up in the capacitor in the form of electric potential energy. This energy is recovered as heat when the capacitor is discharged through a resistance.
Let us consider a capacitor of capacitance $C$ Farad, with a potential difference of $V$ Volt between the plates. The charge $Q$ Is equal to $C V$ Coulomb. There is a charge $+ Q$ On one plate and $- Q$ On the other. In the process of charging, electrons are transferred from the positive to the negative plate, until each plate acquires an amount of charge $Q$ . Suppose during the process of charging, we increase the charge from $Q'$ to $Q' + dQ'$ By transferring an amount of negative charge $dQ'$ From the positive to the negative plate. The potential difference between the plates at that instant will be $V' = Q' / C$ . The work that has to be done is

$dW = V' dQ' = \frac{Q'}{C} dQ' .$

Therefore, the total work done in charging the capacitor from the uncharged state to the final charge $Q$ Will be

$W = \int_{0}^{Q} \frac{Q'}{C} dQ' = \frac{1}{C} {[\frac{Q'^{2}}{2}]}_{0}^{Q} = \frac{1}{2} \frac{Q^{2}}{C} .$

But $Q = C V .$

$∴ W = \frac{1}{2} \frac{Q^{2}}{C} = \frac{1}{2} C V^{2}$ .

This is the energy $U$ Which is “stored” in the capacitor. Thus,

$U = \frac{1}{2} \frac{Q^{2}}{C} = \frac{1}{2} C V^{2} joule$ .

Derive an expression for the energy stored in a charged parallel-plate air capacitor. How does the energy stored change when the medium of air between the plates is replaced by a medium of dielectric constant $K$ Explain.?

Important Questions on Capacitors and Dielectrics

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