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: Recall the formula for the capacitance of a parallel-plate capacitor. The capacitance \(C_0\) of a parallel-plate capacitor with plate area \(A\) and separation \(d\), filled with vacuum, is: \[ C_0 = \frac{\epsilon_0 A}{d} \] where \(\epsilon_0\) is the permittivity of free space.
Step 2: Recall the effect of inserting a dielectric material. When a dielectric material with dielectric constant \(K\) is inserted to completely fill the space between the plates, the permittivity of the space becomes \(\epsilon = K \epsilon_0\).
Step 3: Write the formula for the new capacitance. The new capacitance \(C\) with the dielectric slab is: \[ C = \frac{K \epsilon_0 A}{d} \]
Step 4: Compare the new capacitance with the original capacitance. \[ C = K \left( \frac{\epsilon_0 A}{d} \right) = K C_0 \] Thus, the capacitance increases by a factor of K. Since the dielectric constant \(K\) is always greater than 1 for materials, the capacitance always increases.