Three point charges q, – 4q and 2q are placed at the vertices of an equilateral triangle ABC of side ‘l’ as shown in the figure. Obtain the expression for the magnitude of the resultant electric force acting on the charge q.
(b) Find out the amount of the work done to separate the charges at infinite distance.

Solution

$| {\vec{F}}_{1} | = \frac{1}{4 {πε}_{0}} \frac{(4 q) (q)}{l^{2}} F_{1} = \frac{1}{4 {πε}_{0}} \frac{(4 q^{2})}{l^{2}} F_{1} = \frac{1}{{πε}_{0}} \frac{q^{2}}{l^{2}} {\vec{F}}_{2} | = \frac{1}{4 {πε}_{0}} \frac{(2 q) (q)}{l^{2}} F_{2} = \frac{1}{2 {πε}_{0}} \frac{q^{2}}{l^{2}} angle between \vec{F_{1}} and \vec{F_{2}} is 120^{o} \vec{F} = \sqrt{F_{1}^{1} + F_{2}^{2} + 2 F_{1} F_{2} \cos 120^{o}} F_{1} = 2 F_{2} F = \sqrt{(2 F_{2})^{2} + F_{2}^{2} + 4 F_{2}^{2} \cos 120^{0}} F = \sqrt{4 F_{2}^{2} + F_{2}^{2} - 2 F_{2}^{2}} F = \sqrt{3 F_{2}^{2}} F = \sqrt{3 F_{2}^{2}} F = \sqrt{3} \frac{1}{2 {πε}_{0}} \frac{q^{2}}{l^{2}}$

(b) The amount of work done to separate the charges at infinity will be equal to potential energy.

$U = \frac{1}{4 {πε}_{0} l} [q x (- 4 q) + (q x 2 q) + (- 4 q x 2 q) U = \frac{1}{4 {πε}_{0} l} [- 4 q^{2} + 2 q^{2} - 8 q^{2}] U = \frac{1}{4 {πε}_{0} l} [- 10 q^{2}] U = - \frac{1}{4 {πε}_{0} l} [10 q^{2} unit$

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Electric Charges And Fields

Some More Questions From Electric Charges and Fields Chapter

a) An electrostatic field line is a continuous curve. That is, a field line cannot have sudden breaks. Why not?
b) Explain why two field lines never cross each other at any point?

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