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Question: Two metal plates having charge \( + Q\) and \( - Q\) face each other at a certain distance between t...

Two metal plates having charge +Q + Q and −Q - Q face each other at a certain distance between them. If the plates are now dipped in a kerosene oil tank, the electric fields between the plates will
A) Become zero
B) Increase
C) Decrease
D) Remains zero

Explanation

Solution

In this solution, we will first calculate the formula for electric fields between two metal plates. Then we will determine its dependence on permittivity which will be affected when the metal plates are dipped in kerosene.
Formula used: In this solution, we will use the following formulae:
Electric field between two charged plates E=σε0E = \dfrac{\sigma }{{{\varepsilon _0}}} where σ\sigma is the surface charge density and ε0{\varepsilon _0} is the permittivity in a vacuum.

Complete step by step answer:
We’ve been given that two opposite electrically charged plates are placed parallel to each other. We know that the electric field due to one charged plate is proportional to the charge density on the plate. This can be mathematically written as:
E=σ2ε0E = \dfrac{\sigma }{{2{\varepsilon _0}}}
Since the plates are charged opposite, their electric fields will be in the same direction, and hence the net electric field due to 2 plates add up. So the net electric field will be:
Enet=σ2ε0+σ2ε0{E_{net}} = \dfrac{\sigma }{{2{\varepsilon _0}}} + \dfrac{\sigma }{{2{\varepsilon _0}}}
⇒Enet=σε0\Rightarrow {E_{net}} = \dfrac{\sigma }{{{\varepsilon _0}}}
As we can see in the above equation, the electric field is inversely proportional to the permittivity of the medium in which the two plates are placed in.
Now when the two plates are dipped in kerosene, the permittivity will increase. From the above relation, we can see that as the permittivity increases, the electric field will decrease.

So, option (C) is correct.

Note: We don’t have to know the permittivity of the kerosene to answer this question. Unless mentioned otherwise, we can assume the medium between the plates in the prior case to be a vacuum, and hence the permittivity of any medium will be higher than vacuum so the electric field will be lower for any medium.