Question

On decreasing the charge on both particles by a factor of $3$ while leaving all other factors the same will:

• A. Cause the force to decrease by a factor of $9$.
• B. Cause the force to decrease by a factor of $2$.
• C. Cause the force to increase by a factor of $4$.
• D. Cause the force to increase by a factor of $2$.
• E. Cause the force to remain the same.

Answer 1

Answer: (A)

Cause the force to decrease by a factor of $9$.

Let the initial charges of both the particles be $q_1$ and $q_2$.

Thus initial coulomb force $F=\frac{k{q}_1{q}_2}{r^2}$

Now the charge of both the particles is reduces by a factor of 3, i.e $q_1^{\prime }=\frac{q_1}{3}$ and $q_2^{\prime }=\frac{q_2}{3}$

Thus new coulomb force between them $F=\frac{k\left(q_1/3\right)\times \left(q_2/3\right)}{r^2}=\frac{F}{9}$

Hence the force decreases by a factor of $9$.