Two point charges q_1 = 6 mutext{C} and q_2 = | vedclass

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(A) The electrostatic potential energy of a system of two point charges is given by $U = \frac{1}{4\pi\epsilon_0} \frac{q_1 q_2}{r}$.Initial distance

Vedclass · Accepted Answer

$0.54 	ext{J}$

Vedclass · Answer

(A) The electrostatic potential energy of a system of two point charges is given by $U = \frac{1}{4\pi\epsilon_0} \frac{q_1 q_2}{r}$.Initial distance $r_1 = 10 	ext{cm} = 0.1 	ext{m}$.Final distance $r_2 = 10 	ext{cm} - 2 	ext{cm} = 8 	ext{cm} = 0.08 	ext{m}$.Initial potential energy $U_1 = (9 	imes 10^9) 	imes \frac{(6 	imes 10^{-6}) 	imes (4 	imes 10^{-6})}{0.1} = \frac{9 	imes 24 	imes 10^{-3}}{0.1} = 2.16 	ext{J}$.Final potential energy $U_2 = (9 	imes 10^9) 	imes \frac{(6 	imes 10^{-6}) 	imes (4 	imes 10^{-6})}{0.08} = \frac{9 	imes 24 	imes 10^{-3}}{0.08} = 2.70 	ext{J}$.Increase in potential energy $\Delta U = U_2 - U_1 = 2.70 	ext{J} - 2.16 	ext{J} = 0.54 	ext{J}$.

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