Two charges 7 mu C and -2 mu C are placed a | vedclass

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Question

(C) The potential $V$ due to the external field $E = \frac{A}{r^2}$ is given by $V = -\int_{\infty}^{r} E \cdot dr = -\int_{\infty}^{r} \frac{A}{r^2}

Vedclass · Accepted Answer

$49.3$

Vedclass · Answer

(C) The potential $V$ due to the external field $E = \frac{A}{r^2}$ is given by $V = -\int_{\infty}^{r} E \cdot dr = -\int_{\infty}^{r} \frac{A}{r^2} dr = \frac{A}{r}$.Here, $r_1 = 9 \ cm = 0.09 \ m$ and $r_2 = 9 \ cm = 0.09 \ m$.The total electrostatic potential energy $U$ is the sum of the potential energy of each charge in the external field and their mutual interaction energy:$U = q_1 V(r_1) + q_2 V(r_2) + \frac{k q_1 q_2}{r_{12}}$$U = (7 	imes 10^{-6}) \left( \frac{9 	imes 10^5}{0.09} ight) + (-2 	imes 10^{-6}) \left( \frac{9 	imes 10^5}{0.09} ight) + \frac{(9 	imes 10^9) (7 	imes 10^{-6}) (-2 	imes 10^{-6})}{0.18}$$U = (7 	imes 10^{-6}) (10^7) - (2 	imes 10^{-6}) (10^7) - \frac{126 	imes 10^{-3}}{0.18}$$U = 70 - 20 - 0.7 = 49.3 \ J$.

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