In the rectangle, shown below, the two corners have charges $q_{1} = 5\mu C$ and $q_{2} = + 2.0\mu C.$ The work done in moving a charge $+ 3.0\mu C$ from $B$ to $A$ is (take 1/4 $\pi\varepsilon_{0} = 10^{10}\ N - m^{2}/C^{2}$) [image:1_image170.emf]

Question

Dr. Abubacker · Accepted Answer

(a) Work done $W = 3 	imes 10^{- 6}\left( V_{A} - V_{B} ight);$ where
$V_{A} = 10^{10}\left\lbrack \frac{\left( - 5 	imes 10^{- 6} ight)}{15 	imes 10^{- 2}} + \frac{2 	imes 10^{- 6}}{5 	imes 10^{- 2}} ightbrack = \frac{1}{15} 	imes 10^{6}\ volt$
And $V_{B} = 10^{10}\left\lbrack \frac{\left( 2 	imes 10^{- 6} ight)}{15 	imes 10^{- 2}} + \frac{5 	imes 10^{- 6}}{5 	imes 10^{- 2}} ightbrack = - \frac{13}{15} 	imes 10^{6}\ volt$
$	herefore W = 3 	imes 10^{- 6}\left\lbrack \frac{1}{15} 	imes 10^{6} - \left( - \frac{13}{15} 	imes 10^{6} ight) ightbrack = 2.8\ J$

Q. In the rectangle, shown below, the two corners have charges $q_{1} = 5\mu C$ and $q_{2} = + 2.0\mu C.$ The work done in moving a charge $+ 3.0\mu C$ from $B$ to $A$ is (take 1/4 $\pi\varepsilon_{0} = 10^{10}\ N - m^{2}/C^{2}$)

Explanation

Reviewed by Dr. Abubacker

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