Question

Two point charges having equal charge separated by 1 m distance experience a force of 8 N. What will be the force experienced by them if they are held in water at the same distance. (Given dielectric constant of water ,K water=80)

Answer 1

Ebsilon of vaccum/ebsilon of medium=k So Ebsilon of vaccum= k×ebsilon of medium. Put the value of ebsilon of vaccum in the formula..and u wil get the answer

Answer 2

Force experienced in water is 0.1 N

Given:

Distance between two point charges = 1 m  

Electrostatic force in air = 8 N  

Solution:

Thereby, the electrostatic force is determined by the formula given below:

$F_{\text {air}}=\left(\frac{1}{4 \pi \varepsilon_{\mathrm{o}}}\right) \frac{q 1 q 2}{r^{2}}$

Dielectric constant is given for water to be 80

$K=\frac{\varepsilon_{r}}{\varepsilon_{0}}$

So, force in water is given as,

$F_{\text {water}}=\left(\frac{1}{4 \pi \varepsilon_{r}}\right) \frac{q 1 q 2}{r^{2}}$

Thereby, in water,

$F=\left(\frac{1}{4 \pi K \varepsilon_{0}}\right) \frac{q 1 q 2}{r^{2}}$

Solving both the equation, we get,

$\frac{\mathrm{F}_{\text { air }}}{\mathrm{F}_{\text { water }}}=\mathrm{K}$

$\mathrm{F}_{\text { water }}=\frac{8}{80}$

$\mathrm{F}_{\text { water }}=0.1 \ \mathrm{N}$