The Question is from neet-physics electrostatics , and it was created by the author P.NARASIMHA

Question

A parallel plate capacitor with capacitance C is charged to a potential V. It is then disconnected from the battery and the distance between the plates is doubled. What happens to the energy stored in the capacitor?

P.NARASIMHA · Accepted Answer

## CORE INFORMATION
✓ **Answer**: B
## CONCEPT FRAMEWORK
	**Primary Concept**: Energy stored in a capacitor and its dependence on capacitance and voltage.
	**Related Topics**: Capacitance, potential difference, energy storage, dielectrics.
	**Prerequisites**: Understanding of capacitance, voltage, and energy relationships in capacitors.
## SOLUTION PATHWAY
**Given Information**:
	- Initial capacitance: $C$
	- Initial voltage: $V$
	- Distance between plates is doubled.
**Method & Steps**:
	1. **Capacitance change:**
		Reasoning: Capacitance of a parallel plate capacitor is inversely proportional to the distance between the plates: $C = \epsilon_0 A / d$. Doubling the distance halves the capacitance.
		Formula/Rule: $C' = C/2$
	2. **Charge remains constant:**
		Working: Since the capacitor is disconnected from the battery, the charge $Q$ on the plates remains constant.
		Key Point:  $Q = CV$ initially, and $Q = C'V'$ after the distance change.
	3. **New voltage:**
		Reasoning: Using the charge conservation, $CV = (C/2)V'$, which implies $V' = 2V$.
	4. **Energy change:**
		Reasoning: The energy stored in a capacitor is given by $U = (1/2)CV^2$.
		Working: Initial energy $U = (1/2)CV^2$. Final energy $U' = (1/2)C'V'^2 = (1/2)(C/2)(2V)^2 = CV^2 = 2U$.
		Result: The energy stored doubles.
		Verification: Alternatively, $U = (1/2)QV$. Since $Q$ is constant and $V$ doubles, $U$ also doubles.
## OPTION ANALYSIS
[A] :x: Halves - This would be the case if the voltage remained constant and the capacitance halved. However, the charge remains constant, leading to a doubling of the voltage.
[B] ✓: Doubles -  As shown in the solution pathway, the energy doubles due to the voltage doubling while the capacitance halves.
[C] :x: Remains the same - This would be true if both capacitance and voltage remained unchanged, which is not the case.
[D] :x: Quadruples - This would happen if the capacitance remained constant and the voltage quadrupled, which isn't the scenario here.
## LEARNING AIDS
	**Common Mistakes**:
		1. Forgetting that charge is conserved when the capacitor is disconnected.: This leads to incorrect calculations of the new voltage.
		2. Confusing the formulas for energy stored in a capacitor.: Ensure you use the correct formula based on the given variables (C and V, or Q and V).
	**Quick Tips**:
		1. Remember the relationship between capacitance and distance: $C \propto 1/d$.
		2. When a charged capacitor is disconnected, the charge remains constant.

P.NARASIMHA · Answer

Halves

P.NARASIMHA · Answer

The correct Option is "Doubles"

P.NARASIMHA · Answer

Remains the same

P.NARASIMHA · Answer

Quadruples

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