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Physics-

General

Easy

Question

In the given figure two tiny conducting balls of identical mass m and identical charge q hang from non-conducting threads of equal length L. Assume that $theta$ is so small that $tan invisible function application theta almost equal to sin invisible function application theta$ , then for equilibrium x is equal to

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

The correct answer is: $open parentheses fraction numerator q to the power of 2 end exponent L over denominator 2 pi epsilon subscript 0 end subscript m g end fraction close parentheses to the power of fraction numerator 1 over denominator 3 end fraction end exponent$

In equilibrium F_e = T sinq ....... (i)
mg = T cosq ....... (ii)
$tan invisible function application theta equals fraction numerator F subscript e end subscript over denominator m g end fraction equals fraction numerator q to the power of 2 end exponent over denominator 4 pi epsilon subscript o end subscript x to the power of 2 end exponent cross times m g end fraction$ also $tan invisible function application theta almost equal to sin invisible function application theta equals fraction numerator x divided by 2 over denominator L end fraction$
Hence $fraction numerator x over denominator 2 L end fraction equals fraction numerator q to the power of 2 end exponent over denominator 4 pi epsilon subscript o end subscript x to the power of 2 end exponent cross times m g end fraction$
Þ $x to the power of 3 end exponent equals fraction numerator 2 q to the power of 2 end exponent L over denominator 4 pi epsilon subscript o end subscript m g end fraction$ Þ $x equals open parentheses fraction numerator q to the power of 2 end exponent L over denominator 2 pi epsilon subscript o end subscript m g end fraction close parentheses to the power of 1 divided by 3 end exponent$

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