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Charge Q is given a displacement $\;\overrightarrow{r}=a \hat{i} + b \hat{j}\;$ in an electric field $\;\overrightarrow{E}=E_{1} \hat{i} + E_{2} \hat{j}\;.$ The work done is

$(a)\;Q\;(E_{1}a + E_{2}b)\qquad(b)\;Q\;\sqrt{(E_{1} a)^2+(E_{2} b)^2}\qquad(c)\;Q\;(E_{1}+E_{2})\;\sqrt{a^2+b^2}\qquad(d)\;Q\;\sqrt{E_{1}^2+E_{2}^2}\;\sqrt{a^2+b^2}$

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Answer : (a) $\;Q\;(E_{1}a + E_{2}b)$
Explanation :
$W=Q \overrightarrow{E} . \overrightarrow{r}$
$=Q\;(E_{1} \hat{i}+E_{2} \hat{j})\;.(a \hat{i}+b \hat{j})$
$=Q\;(E_{1} a +E_{2} b)\;.$
answered Feb 6, 2014 by yamini.v

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