Name: Dielectrics in a Capacitor
Uploaded: 2019-07-19
Duration: 6 min 3 s
Description: Let us use our knowledge about dielectrics and capacitors to find out how the use of dielectrics can increase the capacitance of a system. Learn through this...

Question

Each plate of a parallel plate air capacitor has an area $S$ . What amount of work has to be performed by external agent to slowly increase the distance between the plates from $x_{1} to x_{2}$ if:

$(i)$ the charge of the capacitor, which is equal to $q$ is kept constant in the process.

$(ii)$ the voltage across the capacitor, which is equal to $V$ is kept constant in the process.

Accepted Answer

$(i)$ When distance between plates is $x_{1}$ , capacitance is given by $C_{1} = \frac{S \in_{\circ}}{x_{1}}$ , where $S$ is area of each plate.

Similarly for $x_{2}$ , $C_{2} = \frac{S \in_{\circ}}{x_{2}}$

When charge is kept constant, potential energy of capacitor is given by $\frac{q^{2}}{2 C}$ .

When distance between plates is $x_{1}$ , potential energy of capacitor is given by,

$U_{1} = \frac{q^{2}}{2 C_{1}} = \frac{q^{2} x_{1}}{2 S \in_{0}}$

Similarly for plate separation $x_{2}$ , potential energy is given by,

$U_{2} = \frac{q^{2}}{2 C_{2}} = \frac{q^{2} x_{2}}{2 S \in_{0}}$

Therefore, work to be done in increasing plate separation from $x_{1}$ to $S$ is given by work potential energy theorem,

$W = U_{2} - U_{1} = = \frac{q^{2} x_{2}}{2 S \in_{0}} - \frac{q^{2} x_{1}}{2 S \in_{0}} .$

$(ii)$ In second case, voltage is fixed, therefore potential energy is given by $\frac{1}{2} C V^{2}$ .

So, $U_{1} = \frac{1}{2} C_{1} V^{2} = \frac{1}{2} \frac{S \in_{\circ}}{x_{1}} V^{2}$

Similarly $U_{2} = \frac{1}{2} C_{2} V^{2} = \frac{1}{2} \frac{S \in_{\circ}}{x_{2}} V^{2}$

Hence, work to be done is given by,

$W = \frac{1}{2} C_{2} V^{2} - \frac{1}{2} C_{1} V^{2} = \frac{1}{2} \frac{S \in_{\circ}}{x_{2}} V^{2} - \frac{1}{2} \frac{S \in_{\circ}}{x_{1}} V^{2} = \frac{1}{2} (\frac{S \in_{\circ}}{x_{2}} - \frac{S \in_{\circ}}{x_{1}}) V^{2}$

Important Questions on Capacitance

Learn and Prepare for any exam you want !