Question

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

Answer 1

Answer: (C)

decrease

Answer 2

The electric field between two charged parallel plates placed in air is
$E = \frac{\sigma}{\varepsilon_0} = \frac{Q}{\varepsilon_0 A}$ ...(i)
When the parallel plates are immersed in kerosene oil tank of dielectric constant $K$, the electric field between the plates is
$E' =\frac{\sigma}{K \varepsilon_0 } = \frac{Q}{K \varepsilon_0 A} = \frac{E}{K}$ (Using (i))
$\therefore \, \frac{E'}{E} =\frac{1}{K}$
As $K > 1$, so $E' < E$