Question 497: Electrostats & Capacitance

Question

Two identical tennis balls each having mass 'm' and charge 'q' are suspended from a fixed point by threads of length 'l'. What is the equilibrium separation when each thread makes a small angle '$$	heta$$' with the vertical?

Accepted Answer

$$T\cos 	heta  = mg$$
$$T\sin 	heta  = {{k{q^2}} \over {{x^2}}}$$
$$	an 	heta  = {{k{q^2}} \over {{x^2}mg}}$$
$$	an 	heta  \approx \sin 	heta  \approx {x \over {2L}}$$
$${x \over {2L}} = {{K{q^2}} \over {{x^2}mg}}$$
$$x = {\left( {{{{q^2}L} \over {2\pi {\varepsilon _0}mg}}} ight)^{1/3}}$$

Answer

$$x = {\left( {{{{q^2}l} \over {2\pi {\varepsilon _0}mg}}} \right)^{{1 \over 2}}}$$

Answer

$$x = {\left( {{{{q^2}l} \over {2\pi {\varepsilon _0}mg}}} \right)^{{1 \over 3}}}$$

Answer

$$x = {\left( {{{{q^2}{l^2}} \over {2\pi {\varepsilon _0}{m^2}g}}} \right)^{{1 \over 3}}}$$

Answer

$$x = {\left( {{{{q^2}{l^2}} \over {2\pi {\varepsilon _0}{m^2}{g^2}}}} \right)^{{1 \over 3}}}$$

Electrostats & Capacitance

Two identical tennis balls each having mass 'm' and charge 'q' are suspended from a fixed point by threads of length 'l'. What is the equilibrium separation when each thread makes a small angle '$\theta$' with the vertical?

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