Question 495: Electrostatics & Capacitance

Question

A parallel plate capacitor is charged and then isolated. If the separation between the plates is increased, which one of the following statement is NOT correct?

Accepted Answer

Option B is NOT correct.
In a parallel plate capacitor, the charge $Q$ remains constant after it is isolated, as stated in Option A. When the separation $d$ between the plates is increased, the capacitance $C$ of the capacitor is given by the formula:
$$ C = \frac{\varepsilon A}{d} $$
where $\varepsilon$ is the permittivity of the dielectric material between the plates, and $A$ is the area of the plates. Since $C$ is inversely proportional to $d$, increasing the separation decreases the capacitance, as described in Option D.
The potential difference $V$ across the plates is related to charge and capacitance by
$$ V = \frac{Q}{C} $$
Since the charge $Q$ remains constant and the capacitance $C$ decreases when the separation is increased, the potential difference $V$ must increase, which aligns with Option C. Therefore, the statement that the potential difference decreases (Option B) is incorrect.

Answer

Charge remains constant after it is isolated.

Answer

Potential difference across the plates decreases.

Answer

Potential difference across the plates increases.

Answer

Capacitance of capacitor decreases.

Electrostatics & Capacitance

A parallel plate capacitor is charged and then isolated. If the separation between the plates is increased, which one of the following statement is NOT correct?

Choose the correct answer: