A parallel-plate capacitor with plates of area 590 cm2 and is connected across the terminals of...

A parallel-plate capacitor with plates of area 590 cm2 and is connected across the terminals of a battery. After some time has passed, the capacitor is disconnected from the battery. When the plates are then moved 0.48 cm farther apart, the charge on each plate remains constant but the potential difference between the plates increases by 100 V.

(a) What is the magnitude of the charge on each plate?

(b) Do you expect the energy stored in the capacitor to increase, decrease, or remain constant as the plates are moved this way? Explain your answer.
(c) Support your answer to Part (b) by determining the change in stored energy in the capacitor due to the movement of the plates.

Expert Solution

Let the initia; seperation between the plate is then capacitance will be

Initial voltage will be

Potential after the increment in separation

$V_f=V_i+100=\frac{Q(d+0.48\times10^{-2})}{\epsilon_0 A}\\ \Rightarrow \frac{Qd}{\epsilon_0 A}+100=\frac{Q(d+0.48\times10^{-2})}{\epsilon_0 A}\Rightarrow 100=\frac{Q\times0.48\times10^{-2}}{\epsilon_0 A}\\ \Rightarrow Q=\frac{100\times\epsilon_0 A}{0.48\times10^{-2}}=\frac{100\times8.85\times10^{-12}\times 590\times10^{-4}}{0.48\times10^{-2}}=10.878$ nC$$

(b) Energy stored in the capacitor will increase as by separating the plate, external work is needed to be done on the system (Capacitor) and that work will be stored in the form of electrostatic energy.

(c) Initial, final and change in energy in the capacitor will be

$U_i=\frac{1}{2}QV_i\\ U_f=\frac{1}{2}Q\times(V_i+100)\\ \Delta U=U_f-U_i=\frac{1}{2}Q\times(V_i+100)-\frac{1}{2}QV_i=\frac{1}{2}Q\times100=50Q\\ \Delta U=5.439\times10^{-7}$ J$$

genius_generous answered 2 years ago

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