Two positive point charges are held in place, 0.80 m apart. They are then moved so that their electric potential energy doubles. What is the new separation between the charges?

To solve this problem, we can apply the concept of electric potential energy and the relationship between electric potential energy and separation distance.

The electric potential energy (PE) between two point charges can be calculated using the equation:

PE = k * (q1 * q2) / r

Where:
- PE is the electric potential energy
- k is the electrostatic constant (9 x 10^9 N.m^2/C^2)
- q1 and q2 are the magnitudes of the charges
- r is the separation distance between the charges

In this case, we have two positive point charges, so q1 and q2 are both positive.

Let's assume the initial separation distance between the charges is r1 and the final separation distance is r2. Since the electric potential energy doubles, we can write:

2 * PE = k * (q1 * q2) / r2

Divide both sides of the equation by PE to isolate r2:

2 = k * (q1 * q2) / (r2 * PE)

Now, substitute the equation of electric potential energy (PE = k * (q1 * q2) / r1) into the equation:

2 = k * (q1 * q2) / (r2 * (k * (q1 * q2) / r1))

Simplify and rearrange the equation:

2 = r1 / r2

Multiply both sides of the equation by r2:

2 * r2 = r1

Divide both sides of the equation by 2:

r2 = r1 / 2

Therefore, the new separation distance between the charges is half of the initial separation distance.