Question

two similar equal charges are found to repel each other with a force of 16 Newton .when a placed 0.3 m apart in air .what is the strength of a charge? what will be the force between them if the space between them is filled with a substance of dielectric constant 8?

Answer 1

Given:

Similar , equal charges are found to repel each other by a force of 16 N , when placed at a distance of 0.3 m in air.

To find:

• Magnitude of Charges

• Net force experienced when a material of dielectric constant 8 is kept between them.

Calculation:

Let the charges be q.

$\therefore \: force = \dfrac{1}{4\pi\epsilon_{0}} \bigg \{ \dfrac{ {q}^{2} }{ {d}^{2} } \bigg \}$

$= > \: 16 = \dfrac{1}{4\pi\epsilon_{0}} \bigg \{ \dfrac{ {q}^{2} }{ {(0.3)}^{2} } \bigg \}$

$= > \: 16 = 9 \times {10}^{9} \bigg \{ \dfrac{ {q}^{2} }{ {(0.3)}^{2} } \bigg \}$

$= > \: 16 = 9 \times {10}^{9} \bigg \{ \dfrac{ {q}^{2} }{ 9 \times {10}^{ - 2} } \bigg \}$

$= > {q}^{2} = \dfrac{16}{ {10}^{11} }$

$= > q = \sqrt{ \dfrac{16}{ {10}^{11} } }$

$= > q = \dfrac{4}{ {10}^{(5.5)} }$

$= > q = \dfrac{4}{ 316227.7}$

$= > q = 0.126 \times {10}^{ - 6}$

$= > q = 0.126 \: \mu C$

So the magnitude of charge is 0.126 micro-C.

Now , if a material of dielectric constant 8 is placed in the medium between the charges , let the new force be F2

$\therefore \: F2 = \dfrac{initial \: force}{dielectric \: constant}$

$= > \: F2 = \dfrac{16}{8}$

$= > \: F2 = 2 \: N$

So new Electrostatic Force is 2 N.