Question

The repulsive force between two like charges of 1 coulomb each separated by a distance of 1 m in vacuum is equal to​

Answer 1

Given:

• Magnitude of charge = 1 C
• Distance if separation = 1 metre

To find:

• Electrostatic force of repulsion ?

Calculation:

As per Coulomb's Law of Electrostatic Force, we can say :

$\boxed{F = \dfrac{1}{4\pi \epsilon_{0}} \times \dfrac{q1 \times q2}{ {r}^{2} } }$

• $1/4\pi\epsilon_{0}$ (also called Coulomb's Constant) has a value of 9 × 10⁹.

• q refer to charges , and r refers to distance of separation.

• Putting necessary values in SI unit :

$\implies F = (9 \times {10}^{9}) \times \dfrac{1 \times 1}{ {1}^{2} }$

$\implies F = (9 \times {10}^{9}) \times \dfrac{1 }{1}$

$\implies F = 9 \times {10}^{9} \: N$

So, the forces of repulsion is 9 × 10⁹ Newton.