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(A) The work done $W$ in moving a charge $q_0$ from point $B$ to $A$ is given by $W = q_0(V_A - V_B)$.Given: $q_1 = -5 	imes 10^{-6}\,C$,$q_2 = +2.0

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$2.8$

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(A) The work done $W$ in moving a charge $q_0$ from point $B$ to $A$ is given by $W = q_0(V_A - V_B)$.Given: $q_1 = -5 	imes 10^{-6}\,C$,$q_2 = +2.0 	imes 10^{-6}\,C$,$q_0 = +3.0 	imes 10^{-6}\,C$,$k = 10^{10}\,N\cdot m^2/C^2$.From the geometry of the rectangle:At point $A$: distance from $q_1$ is $15\,cm = 0.15\,m$,distance from $q_2$ is $5\,cm = 0.05\,m$.$V_A = k \left( \frac{q_1}{0.15} + \frac{q_2}{0.05} ight) = 10^{10} \left( \frac{-5 	imes 10^{-6}}{0.15} + \frac{2 	imes 10^{-6}}{0.05} ight) = 10^4 \left( -\frac{5}{0.15} + \frac{2}{0.05} ight) = 10^4 \left( -33.33 + 40 ight) = 6.67 	imes 10^4\,V$.At point $B$: distance from $q_1$ is $5\,cm = 0.05\,m$,distance from $q_2$ is $15\,cm = 0.15\,m$.$V_B = k \left( \frac{q_1}{0.05} + \frac{q_2}{0.15} ight) = 10^{10} \left( \frac{-5 	imes 10^{-6}}{0.05} + \frac{2 	imes 10^{-6}}{0.15} ight) = 10^4 \left( -100 + 13.33 ight) = -86.67 	imes 10^4\,V$.$W = 3.0 	imes 10^{-6} 	imes (6.67 	imes 10^4 - (-86.67 	imes 10^4)) = 3.0 	imes 10^{-6} 	imes (93.34 	imes 10^4) = 2.8\,J$.

In the rectangle shown below,two corners have charges $q_1 = -5\,\mu C$ and $q_2 = +2.0\,\mu C$. The work done in moving a charge $q_0 = +3.0\,\mu C$ from $B$ to $A$ is ......... $J$. (Given: $1/4\pi\varepsilon_0 = 10^{10}\,N\cdot m^2/C^2$)

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