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, C = $\frac {\epsilon_o A}{d}$ decreases. As potential of condenser remains same, charge $Q = CV$ decreases.So option 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 or capacitor are moved apart, its capacity $C = \frac {epsilon_o A}{d}$ decreases. Therefore, its potential , $V =\frac {q}{C} increases$