Question

If the plates of a parallel plate capacitor connected to a battery are moved close to each other, then
A. the charge stored in it, increases.
B. the energy stored in it, decreases.
C. its capacitance increases.
D. the ratio of charge to its potential remains the same.
E. the product of charge and voltage increases.
Choose the most appropriate answer from the options given below :

• A. A, B and E only
• B. A, C and E only
• C. B, D and E only
• D. A, B and C only

Answer 1

The Correct Option is B

When the plates of a parallel plate capacitor connected to a battery are moved closer:

A. The stored charge increases.

B. The stored energy decreases.

C. Its capacitance increases.

D. The ratio of charge to potential remains constant.

E. The product of charge and voltage increases.

Select the most appropriate option:

Analysis:

Capacitance \(C\) of a parallel plate capacitor is \(C = \frac{\epsilon_0 A}{d}\), where \(A\) is plate area and \(d\) is the distance between plates. Decreasing \(d\) increases \(C\).

A. The stored charge increases.

With \(V\) constant (from battery) and \(C\) increasing, \(Q = CV\) increases. Thus, A is correct.

B. The stored energy decreases.

Energy \(U = \frac{1}{2}CV^2\). Since \(C\) increases and \(V\) is constant, \(U\) increases. Thus, B is incorrect.

C. Its capacitance increases.

As explained above, decreasing \(d\) increases \(C\). Thus, C is correct.

D. The ratio of charge to potential remains the same.

The ratio \(Q/V = C\). Since \(C\) increases, the ratio increases. Thus, D is incorrect.

E. The product of charge and voltage increases.

Since \(Q\) increases and \(V\) is constant, the product \(QV\) increases. Thus, E is correct.

Conclusion:

Correct statements are A, C, and E.