Question

When a slab of dielectric medium is placed between the plates of a parallel plate capacitor which is connected with a battery, then the charge on plates in comparision with earlier charge:

• A. is less
• B. is same
• C. is more
• D. depends on the dielectric constant

Answer 1

The correct option is C is more

Before the dielectric is inserted, the charge on the plates is given by $$Q=CV$$After the dielectric of dielectric constant $K$ is inserted with battery connected, the charge becomes $$Q^{\prime }=KCV$$ $[$ when a battery is connected to the capacitor, potential difference across its plates will remain constant $]$

As the dielectric constant is always greater than $1$, so $Q^{\prime }>Q$.

Hence , option (c) is correct answer.

Key Concept- The capacitance increases by a factor $K$ with the introduction of a dielectric material completely occupying the space between the plates.