Question

Consider an equilateral triangle $A B C$ of side $2 a$ in the plane of the paper as shown. The centroid of the triangle is $O$ . Equal charges $(Q)$ are fixed at the vertices $A, B$ and $C$ . In what follows consider all motion and situations to be confined the plane of the paper.

(a) A test charge $(q)$ , of same sign as $Q$ , is placed on the median $A D$ at a point at a distance $δ$ below $O$ . Obtain the force $(F)$ felt by the test charge.
(b) Assuming $δ ≪ a$ discuss the motion of the test charge when it is released.
(c) Obtain the force $(F_{D})$ on this test charge if it is placed at the point $D$ as shown in the figure.
(d) In the figure below mark the approximate locations of the equilibrium point(s) for this system. Justify your answer.
(e) Is the equilibrium at $O$ stable or unstable if we displace the test charge in the direction of $O P$ ? The line $P Q$ is parallel to the base $B C$ . Justify your answer.
(f) Consider a rectangle $A B C D$ . Equal charges are fixed at the vertices $A, B, C$ and $D . O$ is the centroid. In the figure below mark the approximate locations of all the neutral points of the system for a test charge with same sign as the charges on the vertices. Dotted lines are drawn for the reference.
(g) How many neutral points are possible for a system in which $N$ charges are placed at the $N$ vertices of a regular $N$ sided polygon ?.

Accepted Answer

Force on charge $q$ due to charge at $A$ , $F_{q A} = \frac{k Q q}{{(\frac{2 a}{\sqrt{3}} + δ)}^{2}} . . . . (1)$

Force on charge $q$ due to charge at $B$ , $F_{q B} = \frac{k Q q}{a^{2} + {(\frac{a}{\sqrt{3}} - δ)}^{2}} . . . . (2)$

Force on charge $q$ due to charge at $C$ , $F_{q C} = \frac{k Q q}{a^{2} + {(\frac{a}{\sqrt{3}} - δ)}^{2}} . . . . (3)$

$(a)$ Net force on the charge $q$ is $O$

$\Rightarrow F_{net} = \frac{k Q q}{{(\frac{2 a}{\sqrt{3}} + δ)}^{2}} - 2 \frac{k Q q}{a^{2} + {(\frac{a}{\sqrt{3}} - δ)}^{2}} (\frac{(\frac{a}{\sqrt{3}} - δ)}{\sqrt{a^{2} + {(\frac{a}{\sqrt{3}} - δ)}^{2}}}) \Rightarrow F_{net} = \frac{k Q q}{{(\frac{2 a}{\sqrt{3}} + δ)}^{2}} - \frac{2 k Q q (\frac{a}{\sqrt{3}} - δ)}{{(a^{2} + {(\frac{a}{\sqrt{3}} - δ)}^{2})}^{\frac{3}{2}}} = k Q q (\frac{1}{{(\frac{2 a}{\sqrt{3}} + δ)}^{2}} - \frac{2 (\frac{a}{\sqrt{3}} - δ)}{{(a^{2} + {(\frac{a}{\sqrt{3}} - δ)}^{2})}^{\frac{3}{2}}})$

$(b)$ If $δ ≪ a$ , the net force equation can be written as by using binomial expansion

$\Rightarrow F_{net} = \frac{k Q q 9 \sqrt{3} δ}{16 a^{3}}$

Direction of force is in opposite direction of displacement of charge from the centroid, which is the equilibrium position.

Hence, motion of charge will be simple harmonic motion.

$(c)$ When test charge is placed at $D$ , the force due to charges at $B$ and $C$ will be equal and in opposite direction to each other. Hence, will cancel each other.

Net force on test charge will be due to charge at $A$ and its magnitude is given as $\Rightarrow F_{net} = \frac{k Q q}{{(\sqrt{3} a)}^{2}} = \frac{k Q q}{3 a^{2}}$

$(d)$ For small $δ$ , force on the test charge is upwards, while for large $δ$ i.e at $D$ , force is downwards. Hence, there must be a neutral point between $O$ and $D$ . By symmetry there will be neutral points on other medians also. In below figure all possible neutral points are expressed with a dot.

$(e)$ From part $(b)$ , we calculated that, the direction of net force on the particle due to displacement of test particle from $O$ is in opposite direction of its displacement. Hence, the equilibrium at $O$ is stable, which tries to bring the displaced charge particle back to its original equilibrium position.

$(f)$ Equilibrium points are indicated by dot in the below diagram.

$(g)$ When the charge is placed at the corner of $N$ sided polygon, neutral points will lie each on the centroid of the polygon as well as on the line from each vertices to the opposite side or vertices.

Hence, total neutral points will be $N + 1$ .

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