Question

Two charged spheres are initially a distance \(d\) apart. The magnitude of the force on each sphere is \(F\). They are moved closer to each other such that the magnitude of the force on each of them is \(9 F .\) By what factor has the distance between the two spheres changed?

Short answer

Answer: The distance between the two charged spheres has changed by a factor of 1/3.

Step-by-step solution

Remember Coulomb's Law

The formula for the force between two point charges \(q_1\) and \(q_2\) separated by a distance \(r\) is given by Coulomb's Law: \(F = k \frac{q_1 * q_2}{r^2}\) where \(k\) is Coulomb's constant.

Set up initial force equation

We are given that the initial force acting on the two charged spheres is \(F\). Using Coulomb's Law, we can write the initial force equation as: \(F = k \frac{q_1 * q_2}{d^2}\)

Set up final force equation

We are given that the final force acting on the two charged spheres is \(9F\). Let's denote the new distance between the spheres as \(d'\). Using Coulomb's Law, we can write the final force equation as: \(9F = k \frac{q_1 * q_2}{d'^2}\)

Divide final force equation by initial force equation

Divide the final force equation by the initial force equation to eliminate the constants \(k\), \(q_1\), and \(q_2\), since they are the same for both force equations. \(\frac{9F}{F} = \frac{k \frac{q_1 * q_2}{d'^2}}{k \frac{q_1 * q_2}{d^2}}\) Simplify: \(9 = \frac{d^2}{d'^2}\)

Solve for the ratio between the initial and final distances

Now, solve for the ratio \(\frac{d'}{d}\): \(\frac{d'}{d} = \sqrt{\frac{1}{9}} = \frac{1}{3}\)

Interpret the result

The distance between the two charged spheres has changed by a factor of \(\frac{1}{3}\), which means the spheres have been moved closer to each other by a factor of 3.