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: Understanding the Concept:
The capacitance of a parallel plate capacitor depends on the area of the plates and the distance between them. Changing the area affects capacitance directly, while changing the separation affects it inversely.

Step 2: Key Formula or Approach:
The capacitance $C$ is given by: \[ C = \frac{\epsilon_0 A}{d} \] where $A$ is the area of the plates and $d$ is the separation distance.

Step 3: Detailed Explanation:
Let the initial area be $A$ and initial separation be $d$. Initial capacitance: \[ C = \frac{\epsilon_0 A}{d} = 50\mu\text{F} \] Now, the area is reduced to half: $A' = \frac{A}{2}$. The separation is also reduced to half: $d' = \frac{d}{2}$. The new capacitance $C'$ is: \[ C' = \frac{\epsilon_0 A'}{d'} = \frac{\epsilon_0 (A/2)}{(d/2)} \] \[ C' = \frac{\epsilon_0 A}{2} \times \frac{2}{d} = \frac{\epsilon_0 A}{d} = C \] Since $C' = C$, the value remains $50\mu\text{F}$.

Step 4: Final Answer:
The new value of the capacitor is 50μF.