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The potential at a point x due to some charges situated on the x- axis is given by $\;V (x) = \large\frac{20}{(x^2-4)} volt\;$. The electric field at x = 4m is given by

$(a)\;(\large\frac{10}{9}) volt/m (\hat{i})\qquad(b)\;(\large\frac{10}{9}) volt/m(-\hat{i}) \qquad(c)\;(\large\frac{5}{3}) volt/m (\hat{i})\qquad(d)\;(-\large\frac{5}{3}) volt/m (\hat{i})$

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Answer : (a) $\;(\large\frac{10}{9}) volt/m (\hat{i})$
Explanation :
$\overrightarrow{E}=-\large\frac{\partial V(x)}{\partial x} \hat{i}=+\large\frac{20\times2x}{(x^2-4)^2}\;\hat{i}$
$\overrightarrow{E} = \large\frac{20\times2\times4}{12\times12}\;\hat{i}$
$\overrightarrow{E}=\large\frac{10}{9}\;\hat{i}\;.$
answered Feb 6, 2014 by yamini.v
 

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