Question

A charge of +3.0 mC is distributed uniformly along the circumference of a circle with a radius of 20 cm. How much external energy is required to bring a charge of 25mC from infinity to the center of the circle?

Answer 1

Answer:

Work done = 4584.9 J

Explanation:

given: q1=3.0 mC = 3.0 × 10⁻³ C, r = 20 cm = 0.20 m, q1 = 34μC = 34 × 10⁻⁶ C

Solution:

Formula for the potential difference at the center of the circle

P.E = K × q1 q2 /r   (Coulomb's constant k= 8.99 × 10⁹ N·m² / C²)

P.E = 8.99 × 10⁹ N·m² / C² × 3.0 × 10⁻³ C × 34 × 10⁻⁶ C /  0.20 m

P.E =  4584.9 J = Work done