Question

The Electrostatic Force between two identical charges $q_0$ separated by a distance $r_0$ is $F_0$. The magnitudes of each charge and the separation distance are tripled. Find the new Electrostatic Force in terms of $F_0$.

• A. $\frac{1}{9}{F}_0$
• B. $\frac{1}{3}{F}_0$
• C. $F_0$
• D. $3F_0$
• E. $9F_0$

Answer 1

Answer: (C)

$F_0$

Let the distance between the charges initially be $r$.

Thus initial electrostatic force $F_o=\frac{k{Q}_1{Q}_2}{r^2}$

New distance between the charges $r^{\prime }=3r$

Charge of each particle is tripled $Q_1^{\prime }=3Q_1$ and $Q_2^{\prime }=3Q_2$

$\therefore$ New electrostatic force $F^{\prime }=\frac{k\left(3Q_1\right)\left(3Q_2\right)}{(3r{)}^2}=F_o$