Question

A parallel plate capacitor \((C)\) is connected to a battery as shown in the figure. Consider two cases: Case-I: Key \(k\) is kept closed and plates of the capacitor are moved apart using an insulating handle. Case-II: Initially key \(k\) is closed for a long time and then opened. Now, the plates of the capacitor are moved apart using an insulating handle.

Identify the correct statements among the following: (A) In Case-I, \(Q\) remains same but \(C\) changes. (B) In Case-II, \(V\) remains same but \(C\) changes. (C) In Case-I, \(V\) remains same and hence \(Q\) changes. (D) In Case-II, \(Q\) remains same and hence \(V\) changes. Choose the correct answer from the options given below:

• (A) and (B) only
• (C) and (D) only
• (B) and (C) only
• (A) and (D) only

Hint:

For capacitor problems: \[ \text{Battery Connected} \Rightarrow V=\text{constant} \] \[ \text{Battery Disconnected} \Rightarrow Q=\text{constant} \] This is the most important rule for variable-capacitance questions.

Answer 1

The Correct Option is B

Step 1:Analyse Case-I (key closed). Since the capacitor remains connected to the battery, \[ V=\text{constant} \] As the plate separation increases, \[ C \downarrow \] Using \[ Q=CV \] \[ Q \downarrow \] Thus, \[ { \begin{array}{c} V \text{ remains constant} Q \text{ changes} \end{array} } \] Hence, statement (C) is true and statement (A) is false.

Step 2: Analyse Case-II (key opened). After opening the key, the capacitor becomes isolated. Therefore, \[ Q=\text{constant} \] As the plate separation increases, \[ C \downarrow \] Using \[ V=\frac{Q}{C} \] \[ V \uparrow \] Thus, \[ { \begin{array}{c} Q \text{ remains constant} V \text{ changes} \end{array} } \] Hence, statement (D) is true and statement (B) is false.

Step 3: Choose the correct statements. The correct statements are \[ (C)\ \text{and}\ (D) \] \[ { (C)\ \text{and}\ (D)\ \text{only} } \] Hence, the correct option is \[ {(B)} \]