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# A dielectric slab is inserted between the plates of a capacitor. The charge on the capacitor is $Q$ and the magnitude of the induced charge on each surface of the dielectric is $Q'$. A) $Q'$ may be larger than $Q$B) $Q'$ must be larger than $Q$C) $Q'$ must be equal to $Q$D) $Q'$ must be smaller than $Q$

Last updated date: 13th Jun 2024
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Hint: When a dielectric is present between the plates of a capacitor, it will modify the electric field that exists between the two plates of the capacitor. Charges will be induced in the dielectric such that it will produce an electric field opposing the original electric field.
Formula used:
-Charge in a dielectric: $Q' = Q\left[ {1 - \dfrac{1}{k}} \right]$ where $Q$ is the charge on the plates and $k$ is the dielectric constant.

Complete step by step answer:
In the absence of a dielectric, a uniform electric field is present between the capacitor plates. However, when a dielectric is present, as mentioned in the hint, an electric field will be induced in it such that it will oppose the external electric field. This is due to the polarization of charges inside the dielectric material. Hence the charge induced in the dielectric will be due to the property of the dielectric. The magnitude of this charge is given as
$Q' = Q\left[ {1 - \dfrac{1}{k}} \right]$
Now the dielectric constant of any medium is greater than one while vacuum has a dielectric constant of 1. Hence the term inside the square bracket will always have a value less than 1.
So, the charge induced in the dielectric will always be less than the charge on the plates of the capacitor.

So, $Q' < Q$. Hence option (D) is correct.

Note: While the vacuum has a dielectric constant of 1, there is no matter in the vacuum to induce any charges, so again no charge will be induced in it. The charges in the dielectric medium will be removed if the capacitor plates do not have any charges either.