Question

Two point charges having equal charges separated by one meter distance experience a force of 8N.what will be the force experienced by them, if they are held in water,at the same distance?give K for water=80

Answer 1

Answer:

Given:

2 equal charges are separated by 1 meter and they experience a force of 8 N. Dielectric constant of water = 80.

To find:

Force experienced when they are kept in water .

Concept:

Electrostatic force is a medium dependent non-contact force . In other words , the force experienced by both the charges also depend on the medium in which they are placed.

The sole reason for such dependence is the Permittivity (ε) of the medium is different from that of air/vacuum.

Calculation:

Electrostatic force in any medium is given by the following relationship ;

$FORCE_{medium} = \dfrac{FORCE_{air}}{Dielectric \: Constant }$

$= > FORCE_{medium} = \dfrac{8}{80}$

$= > FORCE_{medium} = 0.1 \: N$

So final answer :

$\boxed{ \red{ \sf{ \bold{ \large{ FORCE_{medium} = 0.1 \: N }}}}}$

Answer 2

$</p><p>\huge{\tt{\pink{Hello!!! }}}$

$</p><p>\huge{\fbox{\fbox {\rightarrow {\mathfrak {\green {Question \: - }}}}}}$

• Two point charges having equal charges separated by one meter distance experience a force of 8N.

• What will be the force experienced by them, if they are held in water,at the same distance?

• K water = 80.

$</p><p>\huge{\fbox{\fbox {\rightarrow {\mathfrak {\purple {Answer\: - }}}}}}$

$</p><p>\tt{\orange{\boxed{\boxed{\bold{=>FORCE_{medium}\: = \: \frac{Force_{air}}{K}}}}}}$

$</p><p>\tt{\blue{= >\: FORCE_{medium} \:= \dfrac{8}{80} }}$

$</p><p>\boxed{\boxed{ \purple{ \sf{ \bold{ \large{ FORCE_{medium} = 0.1 \: N }}}}}}$