CBSE Class 12 · 2018 · ElectrostaticsShort Answer

Three point charges

q

- 4 q

and

2 q

are placed at the vertices of an equilateral triangle ABC of side

ℓ

as shown in the figure. Obtain
(a) the expression for the magnitude of the resultant electric force acting on the charge

q

, and
(b) the amount of work done to separate the charges at infinite distance.

Solution

(a) On

q

at A, force due to

- 4 q

(attractive, along AB):

F_{1} = \frac{1}{4 π ε _{0}} \frac{4 q ^{2}}{ℓ ^{2}}

force due to

2 q

(repulsive, along CA):

F_{2} = \frac{1}{4 π ε _{0}} \frac{2 q ^{2}}{ℓ ^{2}}

The angle between them is

12 0^{\circ}

F = F_{1}^{2} + F_{2}^{2} + 2 F_{1} F_{2} cos 12 0^{\circ} = \frac{1}{4 π ε _{0}} \frac{q ^{2}}{ℓ ^{2}} 16 + 4 - 8

F = \frac{1}{4 π ε _{0}} \frac{2 3 q ^{2}}{ℓ ^{2}}

(b) Potential energy of the system:

U = \frac{1}{4 π ε _{0} ℓ} [(q) (- 4 q) + (- 4 q) (2 q) + (2 q) (q)] = - \frac{10 q ^{2}}{4 π ε _{0} ℓ}

Work to separate the charges to infinity:

W = - U = \frac{1}{4 π ε _{0}} \frac{10 q ^{2}}{ℓ}

CBSE Class 12 Board · 2018 · Subjective