Question

two point charges having equal charges separated by distance 1m experience a force of 8n. what will be the force experienced by them, if they are held in water, at the same distance ?

Answer 1

F air=1/4πε˳*q1q2/r²

Fwater=1/4πε˳K*q1q2/r²

Hence

F air/Fwater=K

8/Fwater=80 (Given Kwater=80)

Fwater=1/10N

Answer 2

Given that,

Two point charges having equal charges separated by distance 1 m experience a force of 8 N.

To find,

Force experienced by them if they are held in water at the same distance.

Solution,

The force between two charges is given by :

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

For air K = 1

$F=\dfrac{1}{4\pi \epsilon_o}\dfrac{q_1q_2}{r^2}=8\ N$

For water, K = 80

Now force becomes :

$F'=\dfrac{F}{K}\\\\F'=\dfrac{8}{80}\\\\F'=0.1\ N$

If charges are placed at same distance in water new force becomes 0.1 N.

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