Question

Two point charges q1 and q2 are 3.0 cm apart and (q1+ q2) = 20μC. If the force of repulsion between them is 750N, calculate q1 and q2.

Answer 1

given, force of repulsion between q1 and q2 is 750N

seperation between them is 3cm = 0.03m

using Coulomb's law,

F = kq1.q2/r²

⇒750 = (9 × 10^9q1.q2)/(0.03)²

⇒750 = (9 × 10^9 q1.q2)/(9 × 10^-4)

⇒750 = 10¹³ q1.q2

⇒750 × 10^-13 = q1.q2

⇒75 × 10^-12 = q1.q2 . .......(1)

also given, (q1 + q2) = 20μC = 20 × 10^-6 C .....(2)

using formula, (a - b)² = (a + b)² - 4ab

so, (q1 - q2)² = (q1 + q2)² - 4q1.q2

= (20 × 10^-6)² - 4 × 75 × 10^-12

= 400 × 10^-12 - 300 × 10^-12

= 100 × 10^-12

so, q1 - q2 = 10 × 10^-6 = 10μC .......(3)

from equations (2) and (3),

we get, q1 = 15μC and q2 = 5μC