Question

Two identical small balls having charges q1

+1mC and q2

= – 0.33 mC are brought in

contact and them moved apart to a distance r

= 20 cm. Find the force of their interaction.

(a) 50 kN (b) 25 kN

(c) 20 kN (d) 10 kN​

Answer 1

Explanation:

q1 = 1mC = 10^-3 C

q2 = -0.33 mC = -0.33 × 10^-3 C

When brought in contact the charges get distributed equally.

New charge q1 = q2 = q = 1-0.33/2 × 10^-3 = 0.67 × 10^-3 C

Force between charges = F = kq1q2/r^2 = kq^2/r^2 = (9×10^9 × 0.67 × 0.67 × 10^-6)/(20 × 20 × 10^-4 )

F = (4.04 × 10^3)/(400×10^-4) = (404/400)×10^5 = 1.01 × 10^5 = 10×10^4 = 100kN

I think either my calculations might be wrong or the option d is wrong it is either 100 or 10 kN