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

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 :-

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

Answer 2

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

$\tt\to F = k\dfrac{q_1q_2}{r^2} \\\\\tt\to F = 9\times 10^9 \dfrac{50C\times 15C}{(1m)^2} \\\\\tt\to F = 9 \times 750 \times 10^9 N \\\\\tt\to \boxed{\orange{ \tt Force = 675 \times 10^{10} N }}$