Question

A parallel plate capacitor is connected to a battery as shown in figure. Consider two situations.

(i) Key K is kept closed and plates of capacitors are moved apart using insulating handle

(ii) Key K is opened and plates of capacitors are moved apart using insulating handle.

Which of the following statements is correct?

• A. In (i), Q remains same but C changes.
• B. In (ii) V remains same but C changes.
• C. In (i) V remains same and hence Q changes.
• D. In (ii) both Q and V changes.

Answer 1

Answer: (C)

In (i) V remains same and hence Q changes.

Answer 2

: When key K is kept closed, condenser C is charged to potential V. when plates of capacitors are moved

Apart, its capacitance, $\mathrm { C } = \frac { \varepsilon _ { 0 } \mathrm {~A} } { \mathrm {~d} }$ decreases

As potential of condenser remains same, charge decreases. So option (3) is correct.

Once key K is closed, condenser gets charged, Q = CV

Now , if key K is opened, battery is disconnected, no more charging can occur i.e. Q remains same. As plates of capacitor are moved its capacity

$\mathrm { C } = \frac { \varepsilon _ { 0 } \mathrm {~A} } { \mathrm {~d} }$decreases.

Therefore, its potential, $V = \frac { q } { C }$ increases.