Question

Two point charges in air at a distance of 20 cm from each other interact with a certain force at what distance from each other should these charges be placed in oil of relative permittivity 5 to obtain the same force of interaction

Answer 1

The charges should be placed at 90 cm apart.

Explanation:

The electrostatic forces between the charges in air is given as.

F = q1 x q2 / 4 π εo r^2

Here "r" is the distance between them and ε is the permitivity if free space.

As forces are equal than

r^2 ∝ 1 / ε

Thus

r1^2 / r2^2 = ε2 / ε1

The relative permitivity for space is 1 whereas for oil it is 5.

Thus

r2 = √r1^2 x ε1 / ε2 = √ 400 / 5

   = √80 ~ 90 cm

Hence the charges should be placed at 90 cm apart.

Also learn more

The force between two point charges in air is hundred Newton calculate the force if the distance between them is increased by 50% ?

https://brainly.in/question/3509550

Answer 2

The charges must be placed at a distance of 8.94 cm in oil to obtain the same force of interaction when placed in air.

• Let the charges be $Q_{1},Q_{2}$
• Given,

                  r = 20 cm, $e_{r}$ = 5

     Force of interaction in air = force of interaction in oil

                 $F_{air}=F_{oil}$

                 $\frac{1}{4\pi e_{o}}\frac{Q_{1}Q_{2}}{r^{2}}=\frac{1}{4\pi e_{o}e_{r}}\frac{Q_{1}Q_{2}}{d^{2}}$

                  $\frac{1}{r^{2}}=\frac{1}{e_{r}}\frac{1}{d^{2}}$

                  $\frac{1}{400}=\frac{1}{5 d^{2} }$

                  $d^{2}=\frac{400}{5}$

                  $d=\sqrt{80}$

                  d ≅ 8.94 cm

• Note: Use epsilon when taking permittivity not 'e'. I used 'e' as there is no symbol available for permittivity in editor.