Question

Calculate the force of repulsion acting between the two positive charges 50C and 15C when they kept at a distance of 100cm in air.​

Answer 1

Here the two charges 50C and 15C are kept 1pp cm away. Converting Distance to Sl unit we have 1m. Now on using Columb's Law,

- F = k 9192

r2

50C x 15C

(1m)²

- F = 9 x 10°

→ F = 9 x 750 x 10°N

Force = 675 x 101°N

Answer 2

GiveN :-

Two positive charges 50C and 15C .

They are kept 100 cm away .

To FinD:-

The force of repulsion .

SolutioN :-

We can find the force of repulsion using Columb's Law as :-

$\begin{gathered}\sf:\implies\pink{ Force_{(repulsion)}= \dfrac{1}{4\pi \epsilon_0}\bigg(\dfrac{q_1q_2}{r^2} \bigg)}\\\\\sf:\implies F = \dfrac{1}{4\pi \epsilon_0} \dfrac{(50C)(15C)}{(1m)^2} \\\\\sf:\implies F = 9\times 10^9 \times 750 N \\\\\sf:\implies\underset{\blue{\sf Required \ Force }}{\underbrace{\boxed{\pink{\frak{ Force_{(repulsion)}= 675 \times 10^{10} \ Newtons }}}}}\end{gathered}$