A negative charge -q is placed at the midpoin | vedclass

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Question

(A) Let the two positive charges $Q$ be located at $(0, d)$ and $(0, -d)$. The negative charge $-q$ is displaced to $(x, 0)$.The distance $r$ between

Vedclass · Accepted Answer

$F \propto x$

Vedclass · Answer

(A) Let the two positive charges $Q$ be located at $(0, d)$ and $(0, -d)$. The negative charge $-q$ is displaced to $(x, 0)$.The distance $r$ between each positive charge $Q$ and the negative charge $-q$ is $r = \sqrt{x^2 + d^2}$.The magnitude of the electrostatic force exerted by each positive charge on the negative charge is $F_e = \frac{kQq}{r^2}$.The components of these forces along the $y$-axis cancel out due to symmetry. The components along the $x$-axis add up:$F = -2 F_e \cos 	heta$, where $\cos 	heta = \frac{x}{r}$.Substituting the expressions:$F = -2 \left( \frac{kQq}{r^2} ight) \left( \frac{x}{r} ight) = -\frac{2kQqx}{r^3}$.Since $r = (x^2 + d^2)^{1/2}$, we have:$F = -\frac{2kQqx}{(x^2 + d^2)^{3/2}}$.Given the condition $x \ll d$, we can approximate $x^2 + d^2 \approx d^2$ in the denominator:$F \approx -\frac{2kQqx}{(d^2)^{3/2}} = -\frac{2kQqx}{d^3}$.Thus, the magnitude of the force $F$ is directly proportional to the displacement $x$, i.e.,$F \propto x$.

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