Two point electric charges +10^{-8} text{ C} | vedclass

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(D) Let the two charges be $q_1 = +10^{-8} 	ext{ C}$ and $q_2 = -10^{-8} 	ext{ C}$. The distance between them is $d = 0.1 	ext{ m}$.The centre of t

Vedclass · Accepted Answer

$7.2 	imes 10^4 	ext{ NC}^{-1}$

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(D) Let the two charges be $q_1 = +10^{-8} 	ext{ C}$ and $q_2 = -10^{-8} 	ext{ C}$. The distance between them is $d = 0.1 	ext{ m}$.The centre of the line joining the charges is at a distance $r = d/2 = 0.05 	ext{ m}$ from each charge.The electric field due to a point charge is given by $E = \frac{kq}{r^2}$,where $k = 9 	imes 10^9 	ext{ Nm}^2	ext{C}^{-2}$.At the centre,the electric field due to the positive charge $q_1$ points away from it (towards the negative charge),and the electric field due to the negative charge $q_2$ also points towards it.Since both fields are in the same direction,the total electric field $E_{total} = E_1 + E_2$.$E_1 = \frac{k |q_1|}{r^2} = \frac{9 	imes 10^9 	imes 10^{-8}}{(0.05)^2} = \frac{90}{0.0025} = 3.6 	imes 10^4 	ext{ NC}^{-1}$.$E_2 = \frac{k |q_2|}{r^2} = \frac{9 	imes 10^9 	imes 10^{-8}}{(0.05)^2} = 3.6 	imes 10^4 	ext{ NC}^{-1}$.$E_{total} = 3.6 	imes 10^4 + 3.6 	imes 10^4 = 7.2 	imes 10^4 	ext{ NC}^{-1}$.

Two point electric charges $+10^{-8} \text{ C}$ and $-10^{-8} \text{ C}$ are placed $0.1 \text{ m}$ apart. Find the magnitude of the total electric field at the centre of the line joining the two charges.

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