Question

Explain the potential difference between the plates of a charged parallel plate capacitor decreases when its plates are brought closer.

Hint:

For multiple charge systems: \[ U = \sum \frac{k q_i q_j}{r} \] Work done to separate charges to infinity: \[ W = -U \] If system energy is negative → system is bound.

Answer 1

Potential Difference Decreases When Plates are Brought Closer

For an isolated charged parallel plate capacitor, the charge \(Q\) remains constant.

The potential difference across the plates is

\[ V=\frac{Q}{C} \]

The capacitance of a parallel plate capacitor is

\[ C=\frac{\varepsilon_0A}{d} \]

When the plate separation \(d\) decreases, the capacitance \(C\) increases.

Since \(Q\) remains constant,

\[ V=\frac{Q}{C} \]

Therefore, the potential difference decreases.

\[ \boxed{\text{Bringing the plates closer increases capacitance and decreases the potential difference.}} \]