Question

Two equal charges of magnitude 1.1 x 10 -7 C experience an electrostatic force of 4.2 x 10-4 N. How far apart are the centers of the two charges (K=9X9Nm2/C2)

Answer 1

Answer:

0.51 m

Explanation:

From the question,

Applying coulomb's law,

F = kqq'/r²....................... Equation 1

Where F = Electrostatic force, q = magnitude of the first charge, q' = magnitude of the second charge, r = distance between the two charges, k = coulombs constant

making r the subject of the equation

F×r² = kqq'

r² = kqq'/F

r = √(kqq'/F).................. Equation 2

Given: F = 4.2×10⁻⁴N, q = q' = 1.1×10⁻⁷C, k = 9×10⁹ Nm²/C²

Substitute these values into equation 2

r = √(9×10⁹×1.1×10⁻⁷×1.1×10⁻⁷/4.2×10⁻⁴)

r = √(1.1×1.1×9/4.2)(10⁹×10⁻⁷×10⁻⁷×10⁴)]

r = √(2.59×10⁻¹)

r = √(0.259)

r = 0.508

r ≈ 0.51 m