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# Two point charges $\;q_{1}=3\times10^{-6}\;C\;$ and $\;q_{2}=5\times10^{-6}\;C\;$ are located at $\;(3 , 5 ,1)\;$ and $\;(1 , 3 , 2)\;m$ . Find the force on $\;q_{1}\;$ due to $\;q_{2}\;.$

$(a)\;5\times10^{-3}\;(2 \hat{i}+2 \hat{j}-\hat{k})\qquad(b)\;5\times10^{-4}\;(2 \hat{i}+2 \hat{j}-\hat{k})\qquad(c)\;5\times10^{-3}\;(2 \hat{i}-2 \hat{j}+\hat{k})\qquad(d)\;5\times10^{-3}\;(2 \hat{i}+2 \hat{j}+\hat{k})$

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Answer : (a) $\;5\times10^{-3}\;(2 \hat{i}+2 \hat{j}-\hat{k})$
Explanation : $\overrightarrow{F}_{12}=\large\frac{kq_{1}q_{2}\;(\overrightarrow{r}_{1}-\overrightarrow{r}_{2})}{|\overrightarrow{r}_{1}-\overrightarrow{r}_{2}|^3}$
$\overrightarrow{F}_{12}=\large\frac{9\times10^9\times3\times5\times10^{-12}\;(2 \hat{i}+2 \hat{j}-\hat{k})}{27}$
$\overrightarrow{F}_{12}=10^{-2} \hat{i} + 10^{-2} \hat{j}-5\times10^{-3}\;\hat{k}\;.$

answered Jan 30, 2014 by
edited Aug 11, 2014

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