SATHEE: Electrostatics Question 658

Electrostatics Question 658

Question: In the given two tiny conducting balls of identical mass $m$ and identical charge $q$ hang from non-conducting threads of equal length $L$ . Assume that $θ$ is so small that $\tan θ$

$\approx$

$\sin θ$ , then for equilibrium $x$ is equal to

Options:

A) ${(\frac{q^{2} L}{2 π ε_{0} m g})}^{\frac{1}{3}}$

B) ${(\frac{q^{} L^{2}}{2 π ε_{0} m g})}^{\frac{1}{3}}$

C) ${(\frac{q^{2} L^{2}}{4 π ε_{0} m g})}^{\frac{1}{3}}$

D) ${(\frac{q^{2} L^{}}{4 π ε_{0} m g})}^{\frac{1}{3}}$

Show Answer

Answer:

Correct Answer: A

Solution:

[a] In equilibrium $F_{e} = T \sin θ$

$m g = T \cos θ$

$\tan θ = \frac{F_{e}}{m g} = \frac{q^{2}}{4 π ε_{0} x^{2} \times m g}$ Also $\tan θ \approx \sin θ = \frac{x / 2}{L}$ Hence $\frac{x}{2 L} = \frac{q^{2}}{4 π ε_{0} x^{2} \times m g}$

$\Rightarrow x^{3} = \frac{2 q^{2} L}{4 π ε_{0} m g} \Rightarrow x = {(\frac{q^{2} L}{2 π ε_{0} m g})}^{1 / 3}$

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Electrostatics Question 658

Question: In the given two tiny conducting balls of identical mass m and identical charge q hang from non-conducting threads of equal lengthL . Assume that θ is so small thattan⁡θ

Options:

Answer:

Solution:

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Question: In the given two tiny conducting balls of identical mass $m$ and identical charge $q$ hang from non-conducting threads of equal length $L$ . Assume that $θ$ is so small that $\tan θ$