Question

Calculate the force of repulsion between two
charges 1 C and 1 C separated by distance of
1 m in a medium of relative
permittivity 3.​

Answer 1

Correct Question

Calculate the force of repulsion between two charges 1 C and 1 C separated by distance of 1 m in a medium of relative permittivity 3

Solution

Given,

• Both the charges are measured to be 1 C » Q = q = 1 C

• Seperation Distance,r = 1 m

• Free Space of Permittivity = 3

To finD

The repulsive forces between the charges

$\rule{300}{2}$

• Value of Permittivity in vacuum is one

• The force are repulsive because there are like charges

$\rule{300}{2}$

Electric force is given as

$\boxed{ \boxed{ \sf \: F = \dfrac{KQq}{ {r}^{2} } }}$

But,

$\sf \: K = \dfrac{1}{4\pi \: \epsilon} \\ \\ \huge{ \hookrightarrow \: \boxed{ \boxed{ \sf{F = \dfrac{Qq}{4\pi \: \epsilon \: {r}^{2} } }}}}$

$\rule{300}{2}$

(Substituting the values)

$\hookrightarrow \: \sf \: F = \dfrac{1 \times 1}{4 \times 3.14 \times 3 \times {1}^{2} } \\ \\ \hookrightarrow \: \sf \: F = \dfrac{1}{12 \times 3.14} \\ \\ \hookrightarrow \: \boxed{ \boxed{ \sf{F = 0.26 N}}}$

The repulsive forces are of the magnitude 0.26 N

$\rule{300}{2}$

$\rule{300}{2}$

Answer 2

Answer:

$\large\boxed{\sf{3 \times {10}^{9}\;\;N}}$

Explanation:

Let both the equal charges be 'q'

=> q = 1 C

Separation between them, d = 1 m

Relative permittivity, ${\epsilon}_{r} = 3$

Let the force required be 'f'

We know that, by coulomb's law, we have

$\red{f = \dfrac{ 1 }{4\pi {\epsilon}_{0}} \dfrac{ {q}^{2} }{{\epsilon}_{r} {d}^{2} } }$

Putting the respective values, we get,

$= > \sf{f = 9 \times {10}^{9} \times \frac{ {1}^{2} }{3 \times {1}^{2} } } \\ \\ \sf{= > f = 3 \times {10}^{9} }$

Hence require force is $\red{3 \times {10}^{9}\;\;N}$