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.

Answer 2

The capacitance of a parallel plate capacitor follows \( C = \dfrac{\epsilon_0 A}{d} \), where \(A\) is the plate area and \(d\) is the plate separation. Here the linear dimension of the plates is doubled, so since area scales with the square of a linear dimension, the new area becomes \(A^{\prime} = (2)^2 A = 4A\). The separation is increased to 4 times, so \(d' = 4d\). Substituting these directly:

\[ C^{\prime} = \frac{\epsilon_0 (4A)}{4d} = \frac{\epsilon_0 A}{d} = C \]

Checking this result against each option below shows why only one value is consistent with this substitution.

1. 100μF: This would only follow if the area were scaled by a factor of 4 while the separation were left unchanged, but the question explicitly increases the separation too, so this ignores half the given information.
2. 25μF: This would follow if the separation were increased 4 times while the area were mistakenly left unchanged, producing a net decrease rather than the actual unchanged ratio.
3. 50μF: This is exactly what the substitution above produces: since area increases by a factor of 4 (from doubling a linear dimension) and separation also increases by a factor of 4, the two effects cancel and capacitance stays exactly as it was.
4. 200μF: This would require area to increase by a factor of 4 while separation somehow decreased, the opposite of what happens when separation is quadrupled, so it doesn't match the given conditions.

The area-scaling and separation-scaling effects exactly cancel each other out in this problem, leaving the capacitance unchanged.

Therefore, the correct answer is 50μF.