Question

Two spheres carrying charges $+6\mu C$ and $+9\mu C$ separated by a distance $d$, experiences a force of repulsion $F$. When a charge of $-3\mu C$ is given to both the sphere and kept at the same distance as before, the new force of repulsion is :

• A. $F$
• B. $3F$
• C. $F/3$
• D. $F/9$

Answer 1

The correct option is C $F/3$

$\text{Step 1: Coulomb's law in initial situation [Refer Fig. 1]}$

$Q_1=+6\mu C,Q_2=+9\mu C$

$F=\frac{K{Q}_1{Q}_2}{r^2}$ $=\frac{K\left(6\mu C\right)\left(9\mu C\right)}{d^2}$ $....\left(1\right)$

$\text{Step 2: Final charge distribution when}-3\mu C\text{charge added to both the sphere.}$ $\text{[Refer Fig. 2]}$

$\text{Step 3: Apply Coulomb's law in final Situation [Refer Fig. 3]}$

Distance between the spheres remains same $=d$.

$F_1=\frac{K{Q}_1^{\prime }{Q}_2^{\prime }}{d^2}$ $=\frac{K\left(3\mu C\right)\left(6\mu C\right)}{d^2}$ $....\left(2\right)$

Divide equation $\left(2\right)$ by equation $\left(1\right)$

$\frac{F_1}{F}=\frac{6\times 3}{6\times 9}\Rightarrow F_1=\frac{F}{3}$

Hence force in final situation is equal to $\frac{F}{3}$

Option $\left(C\right)$ correct