Four point positive charges of same magnitude | vedclass

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Question

(A) Let the side of the square be $L$. The distance of each charge $Q$ from the centre of the square is $r = \frac{L}{\sqrt{2}}$.When a negative charg

Vedclass · Accepted Answer

The negative charge oscillates along the $Z$-axis.

Vedclass · Answer

(A) Let the side of the square be $L$. The distance of each charge $Q$ from the centre of the square is $r = \frac{L}{\sqrt{2}}$.When a negative charge $-q$ is placed at a distance $z$ from the centre along the $Z$-axis, the distance of the charge from each corner is $d = \sqrt{r^2 + z^2} = \sqrt{\frac{L^2}{2} + z^2}$.The force exerted by each charge $Q$ on $-q$ is $F = \frac{kQq}{d^2}$.The component of this force along the $Z$-axis is $F_z = -F \cos 	heta$, where $\cos 	heta = \frac{z}{d}$.Thus, the net force on the charge $-q$ is $F_{net} = 4 	imes \left( -\frac{kQq}{d^2} 	imes \frac{z}{d} ight) = -\frac{4kQqz}{(\frac{L^2}{2} + z^2)^{3/2}}$.Since $z  Then $F_{net} \approx -\frac{4kQqz}{(L^2/2)^{3/2}} = -\left( \frac{4kQq}{L^3 / 2\sqrt{2}} ight) z = -kz$, where $k$ is a positive constant.Since $F_{net} \propto -z$, the force is a restoring force, and the negative charge will perform simple harmonic motion (oscillate) along the $Z$-axis about the centre of the square.

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