Calculate the force between two charges each of 20 micro coulombs ina

1.	Calculate the force between two charges each of 20 micro coulombs ina medium of relative permittivity 4. The distance between the twocharges is 40 cm
Answer» Question :- Calculate the force between TWO charges each of 20 micro coulombs in a medium of relative permittivity 4. The distance between the two charges is 40 CM . Answer :- $\to \boxed{ \: \: \sf \: F = 5.6 \: N \: \: \: \: } \: \\$ To Find :- → Force between given charges . SOLUTION :- Given that : Charge $\sf q_1 = q_2 = 20 \mu$ distance between charges (R)= 40 cm = 0.4 m relative permittivity $\epsilon_0 = 4$ We know that, _________________ → According to coulomb's law , → Force is directly proportional to the product of charges and inversely proportional to the square of the distance between them. _________________ Hence , $\to \sf \: F = \frac{1}{4 \pi\epsilon_0 \: k } \frac{q_1 q_2}{ {r}^{2} } \\ \\ \because \: 1 \mu = {10}^{ - 6} \\ \\ \to \sf F = \frac{9 \times {10}^{9} }{4 } \: \frac{(20 \times {10}^{ - 6} )(20 \times {10}^{ - 6} )}{ {0.4 \times 0.4}^{ }} \\ \\ \to \sf F = \frac{9 \times 20 \times 20 \times {10}^{ - 1} }{4 \times 4 \times 4} \\ \\ \to \sf \: F = \frac{9 \times 20 \ \: times 2}{4 \times 4 \times 4} \\ \\ \to \sf F = \frac{90}{16} \\ \\ \to \boxed{\sf \: F = 5.6 \: N \: }$ Hence ,force between given charges is 5.6 N

Calculate the force between two charges each of 20 micro coulombs ina medium of relative permittivity 4. The distance between the twocharges is 40 cm

Answer»

Question :-

Calculate the force between TWO charges each of 20 micro coulombs in a medium of relative permittivity 4. The distance between the two

charges is 40 CM .

Answer :-

$\to \boxed{ \: \: \sf \: F = 5.6 \: N \: \: \: \: } \: \\$

To Find :-

→ Force between given charges .

SOLUTION :-

Given that :

Charge $\sf q_1 = q_2 = 20 \mu$

distance between charges (R)= 40 cm = 0.4 m
relative permittivity $\epsilon_0 = 4$

We know that,

_________________

→ According to coulomb's law ,

→ Force is directly proportional to the product of charges and inversely proportional to the square of the distance between them.

_________________

Hence ,

$\to \sf \: F = \frac{1}{4 \pi\epsilon_0 \: k } \frac{q_1 q_2}{ {r}^{2} } \\ \\ \because \: 1 \mu = {10}^{ - 6} \\ \\ \to \sf F = \frac{9 \times {10}^{9} }{4 } \: \frac{(20 \times {10}^{ - 6} )(20 \times {10}^{ - 6} )}{ {0.4 \times 0.4}^{ }} \\ \\ \to \sf F = \frac{9 \times 20 \times 20 \times {10}^{ - 1} }{4 \times 4 \times 4} \\ \\ \to \sf \: F = \frac{9 \times 20 \ \: times 2}{4 \times 4 \times 4} \\ \\ \to \sf F = \frac{90}{16} \\ \\ \to \boxed{\sf \: F = 5.6 \: N \: }$

Hence ,force between given charges is 5.6 N