The distance between charges 5 times 10^{-11} | vedclass

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(C) For a third charge $q$ to experience no net force,the forces exerted by the two charges $Q_1$ and $Q_2$ must be equal in magnitude and opposite in

Vedclass · Accepted Answer

$0.556$

Vedclass · Answer

(C) For a third charge $q$ to experience no net force,the forces exerted by the two charges $Q_1$ and $Q_2$ must be equal in magnitude and opposite in direction. Since the charges have opposite signs,the equilibrium point must lie outside the line segment joining them,closer to the charge with the smaller magnitude $(-2.7 	imes 10^{-11} \, C)$.Let the third charge $q$ be placed at a distance $x$ from the charge $Q_2 = -2.7 	imes 10^{-11} \, C$. The distance from $Q_1 = 5 	imes 10^{-11} \, C$ is $(x + 0.2) \, m$.Equating the magnitudes of the forces: $\frac{k |Q_1| |q|}{(x + 0.2)^2} = \frac{k |Q_2| |q|}{x^2}$.Substituting the values: $\frac{5 	imes 10^{-11}}{(x + 0.2)^2} = \frac{2.7 	imes 10^{-11}}{x^2}$.$\frac{5}{(x + 0.2)^2} = \frac{2.7}{x^2} \implies \frac{\sqrt{5}}{x + 0.2} = \frac{\sqrt{2.7}}{x}$.$2.236 x = 1.643 (x + 0.2) \implies 2.236 x = 1.643 x + 0.3286$.$0.593 x = 0.3286 \implies x \approx 0.554 \, m$. Given the options,the closest value is $0.556 \, m$.

The distance between charges $5 \times 10^{-11} \, C$ and $-2.7 \times 10^{-11} \, C$ is $0.2 \, m$. The distance at which a third charge should be placed in order that it will not experience any force along the line joining the two charges is . . . . . . $m$.

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