There exists a uniform electric field E = 4 t | vedclass

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(A) The potential at the origin is $V_0 = 0 \, V$.The electric field is $E = 4 	imes 10^5 \, Vm^{-1}$ along the negative $x$-axis. The potential at a

Vedclass · Accepted Answer

$120$

Vedclass · Answer

(A) The potential at the origin is $V_0 = 0 \, V$.The electric field is $E = 4 	imes 10^5 \, Vm^{-1}$ along the negative $x$-axis. The potential at a distance $x$ in the direction of the electric field is given by $V = V_0 - E \cdot x$. Since the field is in the negative $x$-direction,the potential increases in the positive $x$-direction: $V(x) = V_0 + E \cdot x$.At $x = 3 \, m$,the potential is $V_3 = 0 + (4 	imes 10^5) 	imes 3 = 12 	imes 10^5 \, V$.The total electrostatic potential energy $U$ of the system is the sum of the interaction energy between the two charges and the potential energy of each charge in the external field:$U = \frac{k q_1 q_2}{r} + q_1 V_0 + q_2 V_3$Given $q_1 = -200 	imes 10^{-6} \, C$,$q_2 = 200 	imes 10^{-6} \, C$,$r = 3 \, m$,$V_0 = 0$,and $V_3 = 12 	imes 10^5 \, V$:$U = \frac{(9 	imes 10^9) 	imes (-200 	imes 10^{-6}) 	imes (200 	imes 10^{-6})}{3} + (-200 	imes 10^{-6})(0) + (200 	imes 10^{-6})(12 	imes 10^5)$$U = \frac{9 	imes 10^9 	imes (-4 	imes 10^{-8})}{3} + 0 + 240$$U = -3 	imes 10^9 	imes 4 	imes 10^{-8} + 240$$U = -120 + 240 = 120 \, J$.

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