A point charge of 50 mu C is placed in the XY | vedclass

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(A) The separation vector $\vec{r}_{sep} = \vec{r} - \vec{r}_0 = (8 \hat{i} - 5 \hat{j}) - (2 \hat{i} + 3 \hat{j}) = 6 \hat{i} - 8 \hat{j} \ m$. The d

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$4.5$

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(A) The separation vector $\vec{r}_{sep} = \vec{r} - \vec{r}_0 = (8 \hat{i} - 5 \hat{j}) - (2 \hat{i} + 3 \hat{j}) = 6 \hat{i} - 8 \hat{j} \ m$. The distance between the points is $r = |\vec{r}_{sep}| = \sqrt{6^2 + (-8)^2} = \sqrt{36 + 64} = \sqrt{100} = 10 \ m$. The magnitude of the electric field is given by $E = \frac{1}{4 \pi \varepsilon_0} \frac{|q|}{r^2}$. Substituting the values: $E = (9 	imes 10^9) 	imes \frac{50 	imes 10^{-6}}{10^2}$. $E = (9 	imes 10^9) 	imes \frac{50 	imes 10^{-6}}{100} = 9 	imes 10^9 	imes 0.5 	imes 10^{-6} = 4.5 	imes 10^3 \ N \ C^{-1}$. Since $1 \ N \ C^{-1} = 1 \ V \ m^{-1}$,we have $E = 4.5 	imes 10^3 \ V \ m^{-1} = 4.5 \ kV \ m^{-1}$.

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