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

for the first one add the force between (q,-4q) and (q,2q). and for the second answer the work done to seperate the chages will be equal to the work done to bring the charges together in equilateral triangle. so the work done is K*2q/l. please check the sign

Answer 2

Answer:

a) The force exerted is $F_{A B}+F_{A C}=\frac{-4 k q^{2}}{l^{2}}+\frac{2 k q^{2}}{l^{2}}=\frac{-2 k q^{2}}{l^{2}}$

b) Work done W = qV

Explanation:

The charges are placed on the side ABC of length l of equilateral triangle as shown:

The force due to charge q will be the sum of electric force exerted on side AB and AC.

Electric force on side AB is

$F_{A B}=\frac{k(q)(-4 q)}{l^{2}}=\frac{-4 k q^{2}}{l^{2}}$

Electric force on side AC is

$F_{A C}=\frac{k(q)(2 q)}{l^{2}}=\frac{2 k q^{2}}{l^{2}}$

So, the force exerted by charge q is $F_{A B}+F_{A C}=\frac{-4 k q^{2}}{l^{2}}+\frac{2 k q^{2}}{l^{2}}=\frac{-2 k q^{2}}{l^{2}}$

The work done to separate the charges placed at infinite distance is:

W=qV, where V is the potential at infinity.