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 :

When the parallel‑plate capacitor remains connected to a battery, the voltage V across the plates is fixed.

The capacitance of a parallel‑plate capacitor is

$C\; =\; \epsilon ₀ A / d$

where ε₀ is the permittivity of free space, A is the plate area and d is the separation. Reducing d (bringing the plates closer) makes the denominator smaller, so C increases.

Because the battery holds V constant, the charge on the plates is

$Q\; =\; C V$

With a larger C, Q must increase. Hence statement A (“the charge stored in it, increases”) is true.

The energy stored in the capacitor is

$U\; =\; ½ C V²$

Since V is unchanged and C grows, U also grows. Therefore statement B (“the energy stored in it, decreases”) is false.

The ratio of charge to potential is

$Q / V\; =\; C$

Because C changes, Q/V does not stay the same. Statement D is false.

The product of charge and voltage is

$Q V\; =\; (C V) V\; =\; C V²$

Again, with C increasing, the product Q V increases. Statement E (“the product of charge and voltage increases”) is true.

Summarizing:

• A – true (charge increases)
• C – true (capacitance increases)
• E – true (Q V increases)
• B – false
• D – false

Thus the most appropriate answer choice is “A, C and E only”.