Two point charges q_1 = 6 mu C and q_2 = 4 mu | vedclass

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(B) The electrostatic potential energy of a system of two point charges is given by $U = \frac{k q_1 q_2}{r}$.Initial separation $r_i = 10 \ cm = 0.1

Vedclass · Accepted Answer

$0.54$

Vedclass · Answer

(B) The electrostatic potential energy of a system of two point charges is given by $U = \frac{k q_1 q_2}{r}$.Initial separation $r_i = 10 \ cm = 0.1 \ m$.Initial potential energy $U_i = \frac{k q_1 q_2}{r_i}$.When $q_2$ is moved towards $q_1$ by $2 \ cm$,the new separation is $r_f = 10 \ cm - 2 \ cm = 8 \ cm = 0.08 \ m$.Final potential energy $U_f = \frac{k q_1 q_2}{r_f}$.The increase in potential energy is $\Delta U = U_f - U_i = k q_1 q_2 \left( \frac{1}{r_f} - \frac{1}{r_i} ight)$.Substituting the values:$k = 9 	imes 10^9 \ N \cdot m^2/C^2$,$q_1 = 6 	imes 10^{-6} \ C$,$q_2 = 4 	imes 10^{-6} \ C$.$\Delta U = (9 	imes 10^9) 	imes (6 	imes 10^{-6}) 	imes (4 	imes 10^{-6}) 	imes \left( \frac{1}{0.08} - \frac{1}{0.1} ight)$.$\Delta U = 216 	imes 10^{-3} 	imes \left( 12.5 - 10 ight)$.$\Delta U = 0.216 	imes 2.5 = 0.54 \ J$.

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