Question

If the air medium between the two charges is placed by water what change you expect in the electrostatic force and why?

Answer 1

see the attachment for the answer

Answer 2

Explanation:

The electric force between charge sis given by :

$F=\dfrac{1}{4\pi \epsilon_o\epsilon_r}\dfrac{q_1q_2}{r^2}$

Here,

$\epsilon_o$ is permittivity of free space

$\epsilon_r$ is relative permittivity

For air, $\epsilon_r=1$

For water, $\epsilon_r=81$

When air is replaced by water force becomes,

$F'=\dfrac{1}{4\pi \epsilon_o81}\dfrac{q_1q_2}{r^2}\\\\F'=\dfrac{1}{81}\times \dfrac{1}{4\pi \epsilon_o}\dfrac{q_1q_2}{r^2}\\\\F'=\dfrac{1}{81}F$

New force gets reduced.

Learn more,

Electric force

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