Two charges each of charge +10 mu C are kept | vedclass

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(C) Let the two charges of $+10 \mu C$ be at $A(0, a)$ and $C(0, -a)$. The charge $-20 \mu C$ is displaced to $B(x, 0)$.The distance between the charg

Vedclass · Accepted Answer

$\frac{3.6 x}{a^3} 	ext{ N}$

Vedclass · Answer

(C) Let the two charges of $+10 \mu C$ be at $A(0, a)$ and $C(0, -a)$. The charge $-20 \mu C$ is displaced to $B(x, 0)$.The distance between the charge at $A$ and the charge at $B$ is $r = \sqrt{a^2 + x^2}$.The electrostatic force $F$ exerted by each charge on the charge at $B$ is given by Coulomb's law:$F = \frac{1}{4 \pi \varepsilon_0} \frac{q_1 q_2}{r^2} = (9 	imes 10^9) \frac{(10 	imes 10^{-6})(20 	imes 10^{-6})}{a^2 + x^2} = \frac{1.8}{a^2 + x^2} 	ext{ N}$.The force $F$ is attractive, directed towards $A$ and $C$. The vertical components of these forces cancel out, while the horizontal components add up.The net force $F_{	ext{net}}$ along the $X$-axis is:$F_{	ext{net}} = 2F \cos 	heta$, where $\cos 	heta = \frac{x}{r} = \frac{x}{\sqrt{a^2 + x^2}}$.Substituting the values:$F_{	ext{net}} = 2 \left( \frac{1.8}{a^2 + x^2} ight) \left( \frac{x}{\sqrt{a^2 + x^2}} ight) = \frac{3.6 x}{(a^2 + x^2)^{3/2}}$.Since $x \ll a$, we can neglect $x^2$ in the denominator:$F_{	ext{net}} \approx \frac{3.6 x}{(a^2)^{3/2}} = \frac{3.6 x}{a^3} 	ext{ N}$.

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