Question

Two small balls having equal positive charge $Q$ (coulomb) on each, are suspended by two insulated string of equal length $L$ metre, from a hook fixed to a stand. The whole set up is taken in satellite into space where there is no gravity (state of weightlessness). Then the angle between. the string and tension in the string is

• A. $180^{\circ },\frac{1}{4\pi {\epsilon }_0}\cdot \frac{Q^2}{(2L{)}^2}$
• B. $75^{\circ },\frac{1}{4\pi {\epsilon }_0}\cdot \frac{Q^2}{L^2}$
• C. $180^{\circ },\frac{1}{4\pi {\epsilon }_0}\cdot \frac{Q^2}{2L^2}$
• D. $180^{\circ },\frac{1}{4\pi {\epsilon }_0}\cdot \frac{QL}{4L^2}$

Answer 1

Answer: (D)

In case of weightlessness, following situation arises

There is a condition of weightlessness in a satellite. Therefore, $mg=0$.
Due to electrostatic force of repulsion between the balls, the string would become horizontal. Therefore, angle between the two small balls $=180^{\circ }$.
From free body diagram, the tension in the string will be equal to the electrostatic force of repulsion,
$<br/>\Rightarrow T=F_e<br/>$
$<br/>\Rightarrow T=\frac{1Q\times Q}{4\pi {ϵ}_o(2L{)}^2}<br/>$
$<br/>\Rightarrow T=\frac{1Q^2}{4\pi {ϵ}_{o}4L^2}N<br/>$