A point charge of 0.009 mu C is placed at th | vedclass

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(B) The electric field $\vec{E}$ due to a point charge $q$ at position vector $\vec{r}$ is given by $\vec{E} = \frac{1}{4\pi \epsilon_0} \frac{q}{r^3}

Vedclass · Accepted Answer

$(3\sqrt{2} \hat{i} + 3\sqrt{7} \hat{j}) \ N C^{-1}$

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(B) The electric field $\vec{E}$ due to a point charge $q$ at position vector $\vec{r}$ is given by $\vec{E} = \frac{1}{4\pi \epsilon_0} \frac{q}{r^3} \vec{r}$.Given $q = 0.009 \ \mu C = 9 	imes 10^{-9} \ C$ and $\vec{r} = \sqrt{2} \hat{i} + \sqrt{7} \hat{j}$.The magnitude of the position vector is $r = \sqrt{(\sqrt{2})^2 + (\sqrt{7})^2} = \sqrt{2 + 7} = \sqrt{9} = 3 \ m$.Substituting the values: $\vec{E} = (9 	imes 10^9) 	imes \frac{9 	imes 10^{-9}}{3^3} (\sqrt{2} \hat{i} + \sqrt{7} \hat{j})$.$\vec{E} = \frac{81}{27} (\sqrt{2} \hat{i} + \sqrt{7} \hat{j}) = 3(\sqrt{2} \hat{i} + \sqrt{7} \hat{j}) = (3\sqrt{2} \hat{i} + 3\sqrt{7} \hat{j}) \ N C^{-1}$.

$A$ point charge of $0.009 \ \mu C$ is placed at the origin. Calculate the electric field intensity due to this point charge at the point $(\sqrt{2}, \sqrt{7}, 0)$.

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