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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 kerosene oil tank, the electric field between the plates will
A
Become zero
B
Increase
C
Decrease
D
Remain same
Detailed Solution
Electric field between the plates in vacuum (air): $E_0 = \frac{\sigma}{\varepsilon_0}$, where $\sigma = \frac{Q}{A}$
The plates are isolated, so the charge Q (and $\sigma$) stays the same when they are dipped in kerosene.
Kerosene is a dielectric with dielectric constant K > 1; it gets polarised and its induced charges set up a field opposing the original field.
Field in the medium: $E = \frac{\sigma}{K\varepsilon_0} = \frac{E_0}{K}$
Since K > 1, $E < E_0$.
Hence the electric field between the plates will decrease.
The plates are isolated, so the charge Q (and $\sigma$) stays the same when they are dipped in kerosene.
Kerosene is a dielectric with dielectric constant K > 1; it gets polarised and its induced charges set up a field opposing the original field.
Field in the medium: $E = \frac{\sigma}{K\varepsilon_0} = \frac{E_0}{K}$
Since K > 1, $E < E_0$.
Hence the electric field between the plates will decrease.