Question

Two identical conducting spheres with negligible volume have $2.1\text{nC}$ and $-0.1\text{nC}$ charges, respectively. They are brought into contact and then separated by a distance of $0.5\text{m}$. The electrostatic force acting between the spheres is

• A. $36\times {10}^{-9}\text{N};$ Attractive Force
• B. $36\times {10}^{-9}\text{N};$ Repulsive Force
• C. $18\times {10}^{-9}\text{N};$ Repulsive Force
• D. $18\times {10}^{-9}\text{N};$ Attractive Force

Answer 1

The correct option is B $36\times {10}^{-9}\text{N};$ Repulsive Force

$q_1=2.1\text{nC}$ and $q_2=-0.1\text{nC}$

The force between them is attractive initially. When they are put in contact, charge sharing takes place between them, it continues to happen until their potential becomes equal. Hence, they will share equal amount of charges.

$q_1^{{}^{\prime }}=q_2^{{}^{\prime }}=\frac{q_1+q_2}{2}=\frac{2.1-0.1}{2}=\frac{2}{2}=1\text{nC}$

Now, the nature of force is repulsive, as like charges repel each other.

Now,

$F=\frac{k{q}_1^{{}^{\prime }}{q}_2^{{}^{\prime }}}{r^2}$

$=\frac{9\times {10}^9\times 1\times {10}^{-9}\times 1\times {10}^{-9}}{{0.5}^2}$

$=36\times {10}^{-9}\text{N};$ Repulsive Force

<!--td {border: 1px solid #ccc;}br {mso-data-placement:same-cell;}--> Hence, $\left(B\right)$ is the correct answer.