Question

Three point charges q, -4q and 2q are placed at the vertices of an equilateral triangle ABC odpf 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 work done to separate the charges at infinite distances.

Answer 1

Answer:

Part a)

Force on charge q due to  two other charges is given as

$F = \frac{2\sqrt3 kq^2}{l^2}$

Part b)

work done to separate these charges is given as

$W = \frac{10 kq^2}{l}$

Explanation:

As we know that the electrostatic force between two charges is given as

$F = \frac{kq_1q_2}{r^2}$

now the force on charge q due to two other charges is given as

$F_1 = \frac{k(q)(-4q)}{l^2} = -2F_o$

$F_2 = \frac{k(q)(2q)}{l^2} = F_o$

so the angle between these two forces is 120 degree here

for resultant for these two forces we have

$F = \sqrt{F_1^2 + F_2^2 + 2F_1F_2cos120}$

$F = \sqrt{4F_o^2 + F_o^2 -(2F_o)(F_o)$

$F = \sqrt3 F_o$

$F = \frac{2\sqrt3 kq^2}{l^2}$

Part b)

As we know that the initial potential energy of system of three charges is given as

$U_i = \frac{k(q)(-4q)}{l} + \frac{k(-4q)(2q)}{l} + \frac{k(q)(2q)}{l}$

$U_i = -\frac{10 kq^2}{l}$

when charges are separated to infinite distance then final potential energy of system of charge is ZERO

so work done to separate these charges is given as

$W = U_f - U_i$

$W = \frac{10 kq^2}{l}$

#Learn

Topic : Electrostatic force and potential energy

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