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. The charge increases
• B. The voltage decreases
• C. The capacitance increases
• D. The electrostatic energy stored increases

Hint:

Remember the golden rule for capacitor transformations: Battery disconnected \(\Rightarrow Q\) remains constant. Battery remains connected \(\Rightarrow V\) remains constant.

Answer 1

The Correct Option is D

Topic of the Question:
The topic of this question is the behavior of a parallel plate capacitor under physical changes, specifically when the plates are separated after being disconnected from a power source.
Step 1 : Understanding the Question:
We are analyzing a parallel plate capacitor that has been fully charged and then completely disconnected from its charging battery. We must determine the effects on charge, voltage, capacitance, and stored electrostatic potential energy when the distance between the plates is increased using insulating handles.
Step 2 : Key Formulas and Approach:

Capacitance of a parallel plate capacitor: $C = \frac{\varepsilon_0 A}{d}$, where $A$ is the area of the plates and $d$ is their separation distance.

Relationship between charge, capacitance, and potential difference: $V = \frac{Q}{C}$.

Electrostatic potential energy stored in the capacitor: $U = \frac{Q^2}{2C}$.

Conservation of charge: When a capacitor is disconnected from a circuit, its plates are electrically isolated, meaning the net charge $Q$ on the plates must remain constant.

Step 3 : Detailed Explanation:

Since the capacitor is disconnected from the battery, there is no conductive path for charge to enter or leave the plates. Therefore, the charge $Q$ remains constant, making Option (A) incorrect.

When the plates are pulled further apart, the separation distance $d$ increases. According to the formula $C = \frac{\varepsilon_0 A}{d}$, the capacitance $C$ is inversely proportional to $d$. An increase in separation leads to a decrease in capacitance, making Option (C) incorrect.

Using the potential relationship $V = \frac{Q}{C}$, we can determine how the voltage changes. Since the charge $Q$ is constant and the capacitance $C$ decreases, the potential difference $V$ across the plates must increase. Thus, Option (B) is incorrect.

Now we evaluate the stored electrostatic potential energy using the formula $U = \frac{Q^2}{2C}$. Since $Q$ is constant and $C$ decreases, the energy $U$ must increase.

This increase in electrostatic energy is physically supplied by the mechanical work performed by an external agent pulling the plates apart against the attractive electrostatic forces between the oppositely charged plates.

Step 4 : Final Answer:
The separation of the plates causes the stored electrostatic energy of the system to increase. This corresponds to the statement in Option (D).