Two parallel metal plates having charges +Q a | vedclass

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(B) The electric field between two parallel plates is given by $E = \frac{\sigma}{\epsilon_0 K}$,where $\sigma$ is the surface charge density,$\epsilo

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Decrease

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(B) The electric field between two parallel plates is given by $E = \frac{\sigma}{\epsilon_0 K}$,where $\sigma$ is the surface charge density,$\epsilon_0$ is the permittivity of free space,and $K$ is the dielectric constant of the medium.When the plates are dipped in kerosene oil,the medium between the plates changes from air (or vacuum) to kerosene oil.The dielectric constant $K$ of kerosene oil is greater than $1$ $(K > 1)$.Since the charge $Q$ on the plates remains constant,the surface charge density $\sigma = \frac{Q}{A}$ also remains constant.Therefore,the new electric field $E'$ becomes $E' = \frac{E}{K}$.Since $K > 1$,the electric field $E'$ will be less than the original electric field $E$. Thus,the electric field decreases.

Two parallel metal plates having charges $+Q$ and $-Q$ face each other at a certain distance between them. If the plates are now dipped in a kerosene oil tank,the electric field between the plates will

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