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An electrical charge $\;4\times10^{-8} C\;$ is placed at the point $\;(2 ,3 ,5)\;.$ At the point $\;(5 , 3, 1) m \;$ electric field vector will be

$(a)\;\large\frac{(216 \hat{i}-288 \hat{k})}{25}\qquad(b)\;\large\frac{(216 \hat{i}+288 \hat{k})}{25}\qquad(c)\;\large\frac{(-216 \hat{i}+288 \hat{k})}{25}\qquad(d)\;\large\frac{(-216 \hat{i}-288 \hat{k})}{25}$

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Answer : (a) $\;\large\frac{(216 \hat{i}-288 \hat{k})}{25}$
Explanation :
$\overrightarrow{E}=\large\frac{k\;q\overrightarrow{r}}{|\overrightarrow{r}|^{3}}$
$\overrightarrow{r}=(5-2) \hat{i} +(3-3) \hat{j} +(1-5) \hat{k}$
$\overrightarrow{r}=3 \hat{i} -4 \hat{k}$
$|\overrightarrow{r}|=5$
$\overrightarrow{E}=\large\frac{9\times10^{9}\times4\times10^{-8}\;(3 \hat{i}-4 \hat{k})}{125}$
$\overrightarrow{E}=\large\frac{216}{25} \hat{i} - \large\frac{288}{25} \hat{k}\;.$
answered Feb 6, 2014 by yamini.v
 

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