Question

The force between two charges is 28 Newton if the paraffin wax of relative permeability 2.8 is introduced between the charges as a medium than the force reduce to?

Answer 1

Answer:

The force between the charges reduces to 10 N.

Explanation:

Given:

The force between the charges initially, $F = 28\ N$.

The relative permeability of the paraffin wax, $\varepsilon_r = 2.8$.

According to Coulomb's law, the electrostatic force between two static point  charges is given by

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

where,

$q_1,\ q_2$ = charges on the two particles.

$r$ = distance between the charges.

$\varepsilon$ = electric permittivity of the medium in which the charges are present.

Initially, when the charges were in vacuum,

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

When paraffin wax is introduced between the charges then the electric force between the charges is given by

$F' =\dfrac{1}{4\pi \varepsilon}\dfrac{q_1q_2}{r^2}......(1)$.

Now, the relative permittivity of the paraffin wax is defined as

$\varepsilon_r = \dfrac{\varepsilon}{\varepsilon_o} \\\varepsilon = \varepsilon_r\varepsilon_o$

Putting this value in expression (1), we get,

$F' =\dfrac{1}{4\pi \varepsilon_r\varepsilon_o}\dfrac{q_1q_2}{r^2}\\=\dfrac{1}{\varepsilon_r}\left(\dfrac{1}{4\pi \varepsilon_o}\dfrac{q_1q_2}{r^2}\right )\\=\dfrac{1}{\varepsilon_r}F\\=\dfrac{1}{2.8}28\\=10\ N.$

So, the force between the charges reduces to 10 N.

Answer 2

Answer:

10N is the correct answer