Question

calculate the force between two charges each of 20 micro coulomb in a medium of relative permittivity 4 .The distance between the two charges is 40 cm

Answer 1

calculate the force between two charges each of 20 micro coulomb in a medium of relative permittivity 4 .The distance between the two charges is 40 cm

Answer 2

Given that,

Charges, $q_1=q_2=20\ \mu C=20\times 10^{-6}\ C$

Relative permittivity, k = 4

The distance between charges, r = 40 cm = 0.4 m

To find,

The force between charges.

Solution,

The force between charges is given by :

$F=\dfrac{1}{4\pi \epsilon_ok}\dfrac{q^2}{r^2}$

$\dfrac{1}{4\pi \epsilon_o}=9\times 10^9$

So,

$F=\dfrac{9\times 10^9}{4}\dfrac{(20\times 10^{-6})^2}{(0.4)^2}\\\\F=5.62\ N$

So, the  force between two charges is 5.62 N.

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