Question

Two charged spheres are \(8.00 \mathrm{~cm}\) apart. They are moved closer to each other by enough that the force on each of them increases four times. How far apart are they now?

Short answer

Answer: 4.00 cm

Step-by-step solution

Recall Coulomb's Law

Coulomb's law is given by the formula: \(F = k * \frac{q_1 * q_2}{r^2}\) Where: \(F\) - the force between the charged spheres \(k\) - Coulomb's constant (\(8.99 × 10^9\) N.m²/C²) \(q_1\) and \(q_2\) - charges of the spheres \(r\) - distance between the center of the charged spheres

Set up the equation for the initial and final situations

Initially, the spheres are \(8.00~cm\) apart, and the force on each of them is \(F_1\). Let's denote the final distance between them as \(r_2\), when the force on each of them increases by a factor of 4 (\(F_2 = 4F_1\)). We can write two equations for each situation: \(F_1 = k * \frac{q_1 * q_2}{r_1^2}\) (initial situation) \(4 * F_1 = k * \frac{q_1 * q_2}{r_2^2}\) (final situation)

Solve the equations for the final distance between the spheres (\(r_2\))

We can rearrange the second equation and substitute the expression from the initial situation equation: \(4 * k * \frac{q_1 * q_2}{r_1^2} = k * \frac{q_1 * q_2}{r_2^2}\) We can divide both sides of the equation by \(k * q_1 * q_2\) to simplify: \(\frac{4}{r_1^2} = \frac{1}{r_2^2}\) Now, we can solve for \(r_2\): \(r_2^2 = \frac{r_1^2}{4}\) \(r_2 = \frac{r_1}{\sqrt{4}}\) \(r_2 = \frac{r_1}{2}\)

Calculate the final distance

Now, we can plug the initial distance (\(r_1 = 8.00~cm\)) into the equation to find the final distance \(r_2\): \(r_2 = \frac{8.00~cm}{2}\) \(r_2 = 4.00~cm\) So, the spheres are now \(4.00~cm\) apart when the force on each of them increases by a factor of four.