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. $180^{\circ}, \frac{1}{4 \pi \varepsilon_{0}} \cdot \frac{Q^{2}}{(2 L)^{2}}$
• B. $75^{\circ}, \frac{1}{4 \pi \varepsilon_{0}} \cdot \frac{Q^{2}}{L^{2}}$
• C. $180^{\circ}, \frac{1}{4 \pi \varepsilon_{0}} \cdot \frac{Q^{2}}{2 L^{2}}$
• D. $180^{\circ}, \frac{1}{4 \pi \varepsilon_{0}} \cdot \frac{Q L}{4 L^{2}}$

Answer 1

Answer: (D)

In case of weightlessness, following situation arises

There is a condition of weightlessness in a satellite. Therefore, $m g=0$.
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,
$
\Rightarrow T = F _{ e }
$
$
\Rightarrow T=\frac{1 Q \times Q}{4 \pi \epsilon_o(2 L)^{2}}
$
$
\Rightarrow T =\frac{1 Q ^{2}}{4 \pi \epsilon_o4 L ^{2}} N
$