Question

An air capacitor C connected to a battery of e.m.f. V acquires a charge q and energy E. The capacitor is disconnected from the battery and a dielectric slab is placed between the plates. Which of the following statements is correct?

• A. V and q decrease but C and E increase
• B. V remains unchange, but q, E and C increase
• C. q remains unchanged, C increases, V and E decrease
• D. q and C increase but V and E decrease

Answer 1

The correct option is C q remains unchanged, C increases, V and E decrease

As the capacitor is disconnected from battery so charge q will remain uncharged.
When dielectric slab placed, the capacitance becomes $C^{\prime }=kC$. where $k$ is dielectric constant. Thus the capacitance C increases k times.
As $V=\frac{q}{C}$, V decreases because q is constant and C increases.
As $V=Ed$, so E also decreases because V decreases.