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Two point charges are located on the x–axis, $q_1 = – 1 m\;C$ at $x = 0$ and $q_2 = + 1 m\;C$ at $x = 1\; m$. Find the work that must be done by an external force to bring a third point charge $q_3 = +1 m\;C$ from infinity to $x = 2$ m.

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Potential at $x = 2$ due to the two charges placed at $x = 0$ and $x = 1$
$V= \large\frac{1}{4 \pi \in_0} \bigg[-\large\frac{1 \times 10^{-6}}{2} + \frac{1 \times 10^{-6}}{1} \bigg]V$
$\quad=4.5 \times 10^3 V$
Hence work done in bringing the charge from infinity will be $W = q_3( \Delta V) $
$\quad= q_3 (V-V_{\infty} = 1 \times 10^{-6} (4.5 \times 10^{3} -0)$
$\quad= 4.5 \times 10^{-3}\;J$
Hence A is the correct answer.
answered Jun 11, 2014 by meena.p
 
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