Electrostatics 1 Question 24
25. A rigid insulated wire frame in the form of a right angled triangle $A B C$, is set in a vertical plane as shown in figure. Two beads of equal masses $m$ each and carrying charges $q_{1}$ and $q_{2}$ are connected by a cord of length $l$ and can slide without friction on the wires.
Considering the case when the beads are stationary determine
(1978)
(a) (i) The angle $\alpha$
(ii) The tension in the cord
(iii) The normal reaction on the beads
(b) If the cord is now cut what are the value of the charges for which the beads continue to remain stationary?
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Answer:
Correct Answer: 25. (a) (i) $\alpha=60^{\circ}$
(ii) $T=\left(\frac{1}{4 \pi \varepsilon_0}\right) \frac{q_1 q_2}{l^2}+m g$
(iii) $N_P=\sqrt{3} m g, N_Q=m g$
(b) $q_1 q_2=-\left(-4 \pi \varepsilon_0\right) m g l^2$
Solution:
- Tension and electrostatic force are in opposite direction and along the string. Now each bead is in equilibrium under three concurrent forces
(i) Normal reaction $(N)$
(ii) Weight $(m g)$ and
(iii) $T-F_{e}$, where $F_{e}=\frac{1}{4 \pi \varepsilon_{0}} \cdot \frac{q_{1} q_{2}}{l^{2}}$
Applying Lami’s theorem for both beads.
$$ \begin{aligned} & \frac{N_{1}}{\sin \left(90^{\circ}-\alpha\right)}=\frac{m g}{\cos \alpha}=\frac{T-F_{e}}{\cos 60^{\circ}} \\ & \frac{N_{2}}{\sin \left(60^{\circ}+\alpha\right)}=\frac{m g}{\sin \alpha}=\frac{T-F_{e}}{\cos 30^{\circ}} \end{aligned} $$
Dividing Eq. (i) by Eq. (ii), we have
$$ \begin{aligned} \tan \alpha & =\frac{\cos 30^{\circ}}{\cos 60^{\circ}}=\sqrt{3} \Rightarrow \alpha=60^{\circ} \\ T & =F_{e}+m g=\frac{1}{4 \pi \varepsilon_{0}} \cdot \frac{q_{1} q_{2}}{l^{2}}+m g \\ N_{1} & =\sqrt{3} m g \quad \text { and } \quad N_{2}=m g \end{aligned} $$
From Eq. (iii) $T=0$ when string is cut.
$$ \text { or } \quad q_{1} q_{2}=-\left(4 \pi \varepsilon_{0}\right) m g l^{2} $$
