Calculate the work done in taking a charge q_ | vedclass

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(B) The electrostatic force is a conservative force,so the work done depends only on the initial and final positions,not on the path taken. The work d

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$1.2$

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(B) The electrostatic force is a conservative force,so the work done depends only on the initial and final positions,not on the path taken. The work done $W$ in moving a charge $q_0$ from $A$ to $B$ is given by $W = q_0(V_B - V_A)$.The potential at a distance $r$ from a point charge $q$ is $V = \frac{kq}{r}$.Given $q = 8 \, mC = 8 	imes 10^{-3} \, C$ at the origin $O(0,0,0)$.Point $A$ is at $(0,0,3 \, cm)$,so $r_A = 3 \, cm = 3 	imes 10^{-2} \, m$.$V_A = \frac{9 	imes 10^9 	imes 8 	imes 10^{-3}}{3 	imes 10^{-2}} = 24 	imes 10^8 \, V$.Point $B$ is at $(0,4 \, cm, 0)$,so $r_B = 4 \, cm = 4 	imes 10^{-2} \, m$.$V_B = \frac{9 	imes 10^9 	imes 8 	imes 10^{-3}}{4 	imes 10^{-2}} = 18 	imes 10^8 \, V$.Work done $W = q_0(V_B - V_A) = (-2 	imes 10^{-9} \, C) 	imes (18 	imes 10^8 - 24 	imes 10^8) \, V$.$W = (-2 	imes 10^{-9}) 	imes (-6 	imes 10^8) \, J = 12 	imes 10^{-1} \, J = 1.2 \, J$.

Calculate the work done in taking a charge $q_0 = -2 \times 10^{-9} \, C$ from point $A$ to point $B$ via point $C$,as shown in the diagram. The source charge $q = 8 \, mC$ is located at the origin $O$. (in $, J$)

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