Question 115: Electrostats & Capacitance

Question

Two charged spheres of $$-20\mu$$C and $$60\mu$$C are kept at a certain distance. They are touched and kept again at the same distance. What is the ratio of force experienced before and after?

Accepted Answer

Given, $$q_1=-20 \mu \mathrm{C}=-2 	imes 10^{-5} \mathrm{C}$$
$$q_2=60 \mu \mathrm{C}=6 	imes 10^{-5} \mathrm{C}$$
Force of attraction between the both charges, kept at a certain distance $$r$$,
$$\begin{aligned}
F_1 & =\frac{K q_1 q_2}{r^2} \
& =\frac{K 	imes 2 	imes 10^{-5} 	imes 6 	imes 10^{-5}}{r^2} \
& =\frac{12 	imes 10^{-10} K}{r^2}
\end{aligned}$$
When both charges are touched, then new charge on both spheres,
$$\begin{aligned}
q_1^{\prime} & =q_2^{\prime}=\frac{q_1+q_2}{2}=\frac{-20+60}{2}=20 \mu \mathrm{C} \
& =2 	imes 10^{-5} \mathrm{C}
\end{aligned}$$
$$	herefore$$ Force between the two spheres,
$$F_2=\frac{K q_1^{\prime} \cdot q_2^{\prime}}{r^2}$$
$$=\frac{K \cdot 2 	imes 10^{-5} 	imes 2 	imes 10^{-5}}{r^2}$$
$$=\frac{4 	imes 10^{-10}}{r^2} K$$
$$	herefore$$ $${{{F_1}} \over {{F_2}}} = {{{{12 	imes {{10}^{ - 10}}K} \over {{r^2}}}} \over {{{4 	imes {{10}^{ - 10}}} \over {{r^2}}}}} = {{12} \over 4} = {3 \over 1}$$
$$	herefore$$ $${F_1}:{F_2} = 3:1$$

Answer

1 : 3

Answer

3 : 1

Answer

2 : 1

Answer

1 : 2

Electrostats & Capacitance

Two charged spheres of $-20\muC and 60\mu$C are kept at a certain distance. They are touched and kept again at the same distance. What is the ratio of force experienced before and after?

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