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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:

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Solution

The correct option is **C** is more

Before the dielectric is inserted, the charge on the plates is given by Q=CVAfter the dielectric of dielectric constant K is inserted with battery connected, the charge becomes Q′=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′>Q.

Hence , option (c) is correct answer.

Before the dielectric is inserted, the charge on the plates is given by Q=CVAfter the dielectric of dielectric constant K is inserted with battery connected, the charge becomes Q′=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′>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. |

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