Question

Part A By what distance do two objects carrying 1.0 C of charge each have to be separated before the electric force exerted on each object is 4.0 N ? Express your answer with the appropriate units.
Part B How much charge should be placed on each of two small objects separated by 1.0 m so that the electric force exerted on each is 4.0 N ? Express your answer with the appropriate units.

Answer 1

Answer:

Part A: 4.74×10⁴ m

Part B: 0.000211 C

Explanation:

Part A:

From coulombs law,

F  = kq²/r²........................ Equation 1

Note: The two object are carrying equal charges.

Where F = Electric force, q = charge, r = distance of separation, k = proportionality constant

Given: F = 4.0 N, q = 1.0 C,

Constant: k = 1/4πe₀ = 9×10⁹ Nm²/C²

Substituting these values into equation 1,

4 = 9×10⁹(1²)/r²

r² = (9×10⁹)/4

r² = 2.25×10⁹

r = √(2.25×10⁹)

r = 4.74×10⁴ m.

Part 2:

F = kq²/r²

Making q the subject of the equation

q = √(Fr²/k)........................... Equation 2

Where: F = 4.0 N, r = 1.0 m, k = 9×10⁹ Nm²/c²

Substituting these values into equation 2

q = √(4×1²/(9×10⁹ )

q = 2.11/10⁴

q = 0.000211 C.