Two point charges 100,mu C and 5,mu C are pla | vedclass

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(D) The work done by an external force in moving a charge $q$ from point $B$ to $C$ in the presence of a source charge $Q$ at $A$ is given by $W = q(V

Vedclass · Accepted Answer

$-2.25$

Vedclass · Answer

(D) The work done by an external force in moving a charge $q$ from point $B$ to $C$ in the presence of a source charge $Q$ at $A$ is given by $W = q(V_C - V_B)$.Here,$Q = 100\,\mu C = 100 	imes 10^{-6}\,C$ and $q = 5\,\mu C = 5 	imes 10^{-6}\,C$.The distance $AB = 40\,cm = 0.4\,m$ and $BC = 30\,cm = 0.3\,m$.Using the Pythagorean theorem,the distance $AC = \sqrt{AB^2 + BC^2} = \sqrt{0.4^2 + 0.3^2} = 0.5\,m$.The potential at $B$ is $V_B = \frac{kQ}{AB} = 9 	imes 10^9 	imes \frac{100 	imes 10^{-6}}{0.4} = 2.25 	imes 10^6\,V$.The potential at $C$ is $V_C = \frac{kQ}{AC} = 9 	imes 10^9 	imes \frac{100 	imes 10^{-6}}{0.5} = 1.8 	imes 10^6\,V$.The work done is $W = q(V_C - V_B) = 5 	imes 10^{-6} 	imes (1.8 	imes 10^6 - 2.25 	imes 10^6) = 5 	imes 10^{-6} 	imes (-0.45 	imes 10^6) = -2.25\,J$.

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