Q136 / 180 · KCET 2015 · Physics
Two spheres carrying charges \( +6 \mu \mathrm{C} \) and \( +9 \mu \mathrm{C} \). separated by a distance \( \mathrm{d} \), experiences a
force of repulsion \( F \). When a charge of \( -3 \mu \mathrm{C} \) is given to both the sphere and kept at the same
distance as before, the new force of repulsion is
force of repulsion \( F \). When a charge of \( -3 \mu \mathrm{C} \) is given to both the sphere and kept at the same
distance as before, the new force of repulsion is
- A\( F \)
- B\( 3 \mathrm{~F} \)
- Correct answerC\( \mathrm{F} / 3 \)
- D\( \mathrm{F} / \mathrm{g} \)
Solution
Given, charge $q_{1}=+6 \mu C=6 \times 10^{-6} C ;$ charge $q_{2}=+9 \mu C$
$=9 \times 10^{-6} C$ seperation $=d$
Force between charges is given as
$F=\frac{1}{4 \Pi \varepsilon_{0}} \frac{q_{1} q_{2}}{d^{2}} \propto \frac{q_{1} q_{2}}{d^{2}}=\frac{6 \times 10^{-6} \times 9 \times 10^{-6}}{d^{2}} \rightarrow(1)$
When a charge of $+3 \mu C$ is given to both and separation $=d$, then
$q_{1}=+6 \mu C-3 \mu C=3 \mu C=3 \times 10^{-6} C$
$q_{2}=+9 \mu C-3 \mu C=6 \mu C=6 \times 10^{-6} C$
The force between charges is given as
$F^{\prime}=\frac{1}{4 \Pi \varepsilon_{0}} \frac{q_{1}^{\prime} q_{2}^{\prime}}{d^{2}} \propto \frac{q_{1}^{\prime} q_{2}^{\prime}}{d^{2}}=\frac{3 \times 10^{-6} \times 6 \times 10^{-6}}{d^{2}} \rightarrow(2)$
Dividing Eq. $(1)$ by Eq. $(2)$, we get
$\frac{F}{F}=\frac{6 \times 10^{-6} \times 9 \times 10^{-6}}{3 \times 10^{-6} \times 6 \times 10^{-6}}=3$
$\Rightarrow F^{\prime}=\frac{F}{3}$ Therefore, the new force of repulsion is $\frac{F}{3}$
Hence, the correct answer is option: 'C'.