Superposition Principle: Overview, Questions, Preparation

An electric charge is the tangible quality of matter that prompts it to experience a force in an electromagnetic field. There are two kinds of electric charges- positive and negative. An object without a net charge is referred to as neutral.

The force on a charge due to other charges is the vector sum of all the forces on that charge of all the other charges, taken one at a time.

Consider an arrangement of three charges q1, q2, and q3. The force on one charge, q1, because of two other charges q2, q3 can be achieved by implementing a vector addition of the forces because of each of these charges. 

I gave the formula for more than three charges as-

F1= F12+ F13+ … + F1n= 1/40[(q1q2/r122)r12+ (q1q3/r132)r13 +... + (q1qn/r1n2)r1n]

F1= q1/40{i=2nq/ir1i2} r1i

The topic- superposition principle is a part of the chapter- electric charges and fields in the class 12 syllabus. This topic carries a weightage of around 3-4 marks in the examination.

1. Consider three charges q1, q2, q3. What is the force on a charge Q placed at the centroid of the triangle in the image below?

Solution:

In the given equilateral triangle ABC of sides of length l, if we draw a perpendicular AD to the side BC, AD = AC cos 30º = ( 3/2 ) l and the distance AO of the centroid O from A is (2/3) AD = (1/ 3 ) l. By symmetry AO = BO = CO.

Thus, Force F1 on Q due to charge q at A is-

F1= (3/40)(Qq/l2)along the line AO

Force F2 on Q due to charge q at B is-

F2=(3/40)(Qq/l2) along the line BO

Force F3 on Q due to charge q at C is-

F3= (3/40)(Qq/l2)along the line CO.

The resultant of forces F2 and F3 is (3/40)(Qq/l2) along the line OA, by the parallelogram law. Therefore, the total force on Q =(3/40)(Qq/l2)(r-r) 0, where r is the unit vector along the line OA.

2. What is the force between two small-charged spheres with charges of 2 × 10−7 C and 3 × 10−7 C placed 30 cm apart in the air?

Solution:

F= q1q2/40r2 = (9x 109x 2x 10-7x 3x 10-7)/(0.3)2= 6x 10-3 N

3. The electrostatic force on a small sphere of charge 0.4 µC due to another small sphere of charge –0.8 µC in the air is 0.2 N. Find the distance between them?

Solution:

The distance r is given by-

r2 = (1/40) q1q2/F

r= 144 x 10-4= 0.12m

Q: What is Coulomb’s Law?

Q: What is meant by an electric field?

Q: What are the inferences from the superposition principle?

Q: Is an electric charge a scalar or a vector quantity?

Q: What does it mean if the total charge is zero?