Two charges 10 mu C and -10 mu C are placed a | vedclass

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(D) The charges are $q_1 = 10 \mu C$ and $q_2 = -10 \mu C$. The distance between them is $2a = 10 \ cm$,so $a = 5 \ cm = 0.05 \ m$.The point $P$ is on

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$4.1 	imes 10^6 \ N \ C^{-1}$

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(D) The charges are $q_1 = 10 \mu C$ and $q_2 = -10 \mu C$. The distance between them is $2a = 10 \ cm$,so $a = 5 \ cm = 0.05 \ m$.The point $P$ is on the perpendicular bisector at a distance $r = 12 \ cm = 0.12 \ m$ from the midpoint.The distance from each charge to point $P$ is $d = \sqrt{r^2 + a^2} = \sqrt{12^2 + 5^2} = \sqrt{144 + 25} = \sqrt{169} = 13 \ cm = 0.13 \ m$.The electric field due to each charge at $P$ has a magnitude $E = \frac{k|q|}{d^2} = \frac{9 	imes 10^9 	imes 10 	imes 10^{-6}}{(0.13)^2} = \frac{9 	imes 10^4}{0.0169} \approx 5.325 	imes 10^6 \ N \ C^{-1}$.The components of the electric fields perpendicular to the line $AB$ cancel out,while the components parallel to $AB$ add up.The net electric field is $E_{	ext{net}} = 2E \cos(	heta)$,where $\cos(	heta) = \frac{a}{d} = \frac{5}{13}$.$E_{	ext{net}} = 2 	imes \left( \frac{9 	imes 10^9 	imes 10 	imes 10^{-6}}{(0.13)^2} ight) 	imes \frac{5}{13} = 2 	imes \frac{9 	imes 10^4}{0.0169} 	imes \frac{5}{13} \approx 4.1 	imes 10^6 \ N \ C^{-1}$.

Two charges $10 \mu C$ and $-10 \mu C$ are placed at points $A$ and $B$ separated by a distance of $10 \ cm$. Find the electric field at a point $P$ on the perpendicular bisector of $AB$,at a distance of $12 \ cm$ from its midpoint.

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