Two parallel metal plates having charges $+Q$ and $-Q$ face each other at a certain distance between them.…
- become zero
- increase
- decrease
- remain same
Solution
The electric field in vacuum
$\mathrm{E}_0=\frac{\sigma}{2 \varepsilon_0}$
When the plates are dipped in kerosene oil tank, then the electric field between the plate will be
$\begin{array}{rlrl}
& & \mathrm{E} & =\frac{\sigma}{2 \varepsilon_0 \mathrm{k}} \\
\because & \mathrm{k} & > 1 \\
\therefore & & \mathrm{E} & < \mathrm{E}_0
\end{array}$
So, the electric field between the plates will decrease.
.Asked in: NEET 2010 (Mains)