Question

According to Coulomb's Law, if the distance between the nucleus and electron were doubled, the force would be__. A. 1/4 as much.. . B. 1/2 as much.. . C. two times as much.. . D. four times as much.

Answer 1

The answer is A. 1/4 as much.

To calculate this, we will use Coulomb's Law:

F = k*Q1*Q2/r²

where F is repulsive force, k is constant, Q is charge, r is distance between charges.

k = 5.0 × 10⁹  N*m/C²

Let's take one example where:

Q1 = Q2 = 1.2 × 10⁻⁶ C²

r₁ = 0.5 m

So, F1 = k*Q1*Q2/r₁² = 5.0 × 10⁹ * 1.2 × 10⁻⁶ * 1.2 × 10⁻⁶ /0.5² = 12.96 × 10⁻²/0.25 = 51.84 × 10⁻² = 0.5184 ≈ 0.52

Let's now double r:

r₂ = 2r₁ = 2*0.5 = 1

And let's calculate force:

F2 = k*Q1*Q2/r₂² = 5.0 × 10⁹ * 1.2 × 10⁻⁶ * 1.2 × 10⁻⁶ /1² = 12.96 × 10⁻²/1 = 12.96 × 10⁻² = 0.1296 ≈ 0.13

And now, let's see how much is the second force (F2) differ from the first force (F1):
F2/F1 = 0.13/0.52 = 1/4