SATHEE: Electrostatics 1 Question 22

Electrostatics 1 Question 22

23. Three particles, each of mass $1 g$ and carrying a charge $q$ , are suspended from a common point by insulated massless strings, each $100 cm$ long. If the particles are in equilibrium and are located at the corners of an equilateral triangle of side length $3 cm$ , calculate the charge $q$ on each particle.

(Take $g = 10 m / s^{2}$ ).

(1988, 5M)

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Answer:

Correct Answer: 23. $3.17 \times 10^{- 9} C$

Solution:

$F$ is the resultant of electrostatic forces between two charges.

$\begin{aligned} F & = 2 F_{e} \cos 30^{\circ} \\ = 2 \frac{1}{4 π ε_{0}} \cdot \frac{q^{2}}{a^{2}} \frac{\sqrt{3}}{2} \\ = \frac{2 \times 10^{9} \times 9 \times q^{2} \times \sqrt{3}}{{(3 \times 10^{- 2})}^{2} \times 2} \\ = \sqrt{3} \times 10^{13} q^{2} \end{aligned}$

$θ$ is the angle of string with horizontal in equilibrium,

$\begin{aligned} \cos θ = \frac{r}{l} = \frac{(a / 2) \sec 30^{\circ}}{l} = \frac{a}{\sqrt{3} l} = \frac{3}{100 \sqrt{3}} \\ ∴ θ \approx 89^{\circ} \end{aligned}$

Now, the particle is in equilibrium under three concurrent forces, $F, T$ and $m g$ . Therefore, applying Lami’s theorem

$\frac{F}{\sin (90^{\circ} + θ)} = \frac{m g}{\sin (180^{\circ} - θ)}$

$\sqrt{3} \times 10^{13} q^{2} = (1 \times 10^{- 3}) (10) \cot 89^{\circ}$

Solving this equation, we get

$q = 0.317 \times 10^{- 8} C or q = 3.17 \times 10^{- 9} C$

Electrostatics 1 Question 22

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Electrostatics 1 Question 22

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