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?

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