Question

A parallel-plate capacitor has a dielectric slab of thickness d and dielectric constant K inserted between the plates. The capacitance change compared to the vacuum case (when no slab is inserted) is:

• A. Increases by a factor of K
• B. Decreases by a factor of K
• C. Increases by a factor of K\(^2\)
• D. Decreases by a factor of K\(^2\)

Hint:

A dielectric material reduces the electric field between the capacitor plates for a given charge. This allows more charge to be stored at the same potential difference, thereby increasing the capacitance (\(C=Q/V\)).

Answer 1

The Correct Option is A

Step 1: State the capacitance formula for a parallel-plate capacitor in a vacuum. The capacitance \(C_0\) is given by \( C_0 = \frac{\epsilon_0 A}{d} \), where \(A\) is the plate area, \(d\) is the separation, and \(\epsilon_0\) is the permittivity of free space.
Step 2: Describe the effect of a dielectric on permittivity. Inserting a dielectric with constant \(K\) changes the permittivity to \(\epsilon = K \epsilon_0\).
Step 3: Formulate the capacitance with the dielectric. The new capacitance \(C\) becomes \( C = \frac{K \epsilon_0 A}{d} \).
Step 4: Compare the capacitances. \( C = K \left( \frac{\epsilon_0 A}{d} \right) = K C_0 \). Since \(K>1\), the capacitance increases by a factor of \(K\).