Question

A charge of $-30\mu C$ is placed at 50 cm from another charge of $+25\mu C$ is vacuum. The force between them is:

• A. $60N$
• B. $27N$
• C. $90N$
• D. $120N$

Answer 1

Answer: (B)

$27N$

$q_1=-30\mu C$
$q_2=25\mu C$
$d=50cm$ or $0.5m$

$\therefore F=\frac{9\times {10}^9\times \left(30\right)\times \left(25\right)\times {10}^{-12}}{(0.5{)}^2}$

$=\frac{9\times 750\times {10}^{-3}}{0.25}=\frac{9\times 75}{25}=27N$

this force will be attractive in nature