Question:
A parallel plate capacitor is charged and then disconnected from the charging battery. If the plates are now pulled further apart using insulating handles:
A parallel plate capacitor is charged and then disconnected from the charging battery. If the plates are now pulled further apart using insulating handles:
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Remember the golden rule for capacitor transformations:
Battery disconnected \(\Rightarrow Q\) remains constant.
Battery remains connected \(\Rightarrow V\) remains constant.
Updated On: May 25, 2026
- The charge increases
- The voltage decreases
- The capacitance increases
- The electrostatic energy stored increases
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The Correct Option is D
Solution and Explanation
Concept:
When a capacitor is disconnected from its charging battery, the isolated system can no longer exchange charge with any source. Therefore, the charge \(Q\) remains constant.
The capacitance \(C\), potential difference \(V\), and electrostatic energy stored \(U\) depend on the geometric configuration formulas:
• Capacitance: \( C = \frac{\varepsilon_0 A}{d} \)
• Potential Difference: \( V = \frac{Q}{C} \)
• Electrostatic Energy: \( U = \frac{Q^2}{2C} \) Step 1: Analyzing the effect of pulling the plates further apart.
When the plates are pulled further apart, the separation distance \(d\) between them increases. From the capacitance relation: \[ C \propto \frac{1}{d} \quad \Rightarrow \quad \text{As } d \text{ increases, } C \text{ decreases.} \] Hence, Option (C) is incorrect.
Step 2: Evaluating the change in voltage and stored energy.
Since the capacitor is disconnected from the battery, the charge \(Q\) must stay constant (\(Q = \text{constant}\)). Hence, Option (A) is incorrect. Using the potential difference relationship: \[ V = \frac{Q}{C} \quad \Rightarrow \quad \text{Since } C \text{ decreases, } V \text{ must increase.} \] Hence, Option (B) is incorrect. Now, evaluating the stored electrostatic energy \(U\): \[ U = \frac{Q^2}{2C} \quad \Rightarrow \quad \text{Since } Q \text{ is constant and } C \text{ decreases, } U \text{ must increase.} \] This additional stored energy comes directly from the mechanical work done by an external agent against the attractive electrostatic forces between the oppositely charged plates.
• Capacitance: \( C = \frac{\varepsilon_0 A}{d} \)
• Potential Difference: \( V = \frac{Q}{C} \)
• Electrostatic Energy: \( U = \frac{Q^2}{2C} \) Step 1: Analyzing the effect of pulling the plates further apart.
When the plates are pulled further apart, the separation distance \(d\) between them increases. From the capacitance relation: \[ C \propto \frac{1}{d} \quad \Rightarrow \quad \text{As } d \text{ increases, } C \text{ decreases.} \] Hence, Option (C) is incorrect.
Step 2: Evaluating the change in voltage and stored energy.
Since the capacitor is disconnected from the battery, the charge \(Q\) must stay constant (\(Q = \text{constant}\)). Hence, Option (A) is incorrect. Using the potential difference relationship: \[ V = \frac{Q}{C} \quad \Rightarrow \quad \text{Since } C \text{ decreases, } V \text{ must increase.} \] Hence, Option (B) is incorrect. Now, evaluating the stored electrostatic energy \(U\): \[ U = \frac{Q^2}{2C} \quad \Rightarrow \quad \text{Since } Q \text{ is constant and } C \text{ decreases, } U \text{ must increase.} \] This additional stored energy comes directly from the mechanical work done by an external agent against the attractive electrostatic forces between the oppositely charged plates.
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