Question

Two point charges of one coulomb separated by 1 metre distance experience byforce of 9 into 10 raise to the power 9 Newton how much force is experienced by them if they emerged in water keeping the distance of separation between them is same dielectric constant for water is equal to 80

Answer 1

Given:

Two point charges of one coulomb separated by 1 m distance experience by a force of 9 × 10^(9) N. Dielectric constant of water is 80.

To find:

New force if they kept in water with same distance of separation.

Calculation:

In air, let force be f1.

$\therefore \: f1 = \dfrac{k(q1)(q2)}{ {d}^{2} }$

$= > \: f1 = \dfrac{k \times 1 \times 1}{ {(1)}^{2} }$

$= > \: f1 = k = 9 \times {10}^{9}$

In water, let force be f2. The dielectric constant is denoted as $\epsilon$.

$\therefore \: f2= \dfrac{k(q1)(q2)}{ \epsilon {d}^{2} }$

$= > \: f2 = \dfrac{k \times 1 \times 1}{ \epsilon{(1)}^{2} }$

$= > \: f2 = \dfrac{k }{ \epsilon}$

$= > \: f2 = \dfrac{9 \times {10}^{9} }{ 80}$

$= > \: f2 = 1.125 \times {10}^{9} \: N$

So, final answer is:

$\boxed{ \red{ \bold{ \: f2 = 1.125 \times {10}^{9} \: N }}}$