Question

The capacitance of a parallel plate capacitor is 50μF. If the linear dimension of the plates are doubled and the separation between the plates is increased to 4 times, what would be the new value of the capacitor?

• A. 100μF
• B. 25μF
• C. 50μF
• D. 200μF

Hint:

Linear dimensions refer to length and width. If linear dimensions are scaled by $k$, the area $A$ scales by $k^2$. Here, $2^2 = 4$, which exactly cancels the 4-fold increase in distance.

Answer 1

The Correct Option is C

Step 1: Capacitance of a parallel plate capacitor follows \( C \propto \dfrac{A}{d} \), so any change in capacitance depends only on how the ratio \(A/d\) changes.

Step 2: The linear dimension of the plates doubles, and since area scales as the square of a linear dimension, the area increases by a factor of \(2^2 = 4\).

Step 3: The separation is increased to 4 times its original value, so \(d\) also increases by a factor of 4.

Step 4: The ratio \(A/d\) changes by a factor of \(4/4 = 1\), meaning it stays exactly the same, so the capacitance is unchanged at its original value.
\[ C' = C = \boxed{50\ \mu\text{F}} \]