Question

The force between two identical charges placed in vacuum separated by 1 cm is equal to 90 N.

What is the magnitude of the two charges?​

Answer 1

Given:

The force between two identical charges placed in vacuum separated by 1 cm is equal to 90 N.

To find:

Magnitude of charges ?

Calculation:

Let the charges be q and the separation distance be d :

• As per COULOMB'S LAW OF ELECTROSTATIC FORCE :

$\boxed{F = \dfrac{1}{4\pi \epsilon_{0}} \times \dfrac{q \times q}{ {d}^{2} }}$

• Now, value of $1/4\pi\epsilon_{0}$ is 9 × 10⁹.

$\implies \: 90 = 9 \times {10}^{9} \times \dfrac{ {q}^{2} }{ {(0.01)}^{2} }$

$\implies \: 90 = 9 \times {10}^{9} \times \dfrac{ {q}^{2} }{ {10}^{ - 4} }$

$\implies \: 90 = 9 \times {10}^{9 + 4} \times {q}^{2}$

$\implies \: 90 = 9 \times {10}^{13} \times {q}^{2}$

$\implies \: 10 = {10}^{13} \times {q}^{2}$

$\implies \: {q}^{2} = {10}^{ - 12}$

$\implies \:q = {10}^{ - 6} \: C$

$\implies \:q = 1 \: \mu C$

So, magnitude of charge is 1 micro-C.