1. An electric charge moves through an electric field. it accelerates, gaining 100J of energy. The change in electric potential of +10V. What is the magnitude of the charge of the particle?

I did 100J/10V to get 10C.

2. If a negative charge of double the magnitude of the first charge moved between the same two points in the field, what is the change in electric potential?

I think it was the same voltage regardless. 10V

3. What is the change in electric potential energy based off of question 2?

Not sure.

Can anyone point out anything that is wrong and show how to do it?

1 and 2 are correct. The charge times the voltage is the change in potential energy. 100 Joules for the 10 Coulombs would be 200 Joules if the charge is doubled.

To answer question 1 correctly:

The change in electric potential (ΔV) is given as +10V, and the energy gained (ΔE) is 100J. The formula relating electric potential difference and energy is ΔE = qΔV, where q represents the charge.

To find the magnitude of the charge, you can rearrange the formula as follows:

q = ΔE / ΔV

Plugging in the given values:

q = 100J / 10V

Simplifying the calculation:

q = 10C

Great job, you correctly calculated that the magnitude of the charge is 10C.

Now let's move on to question 2 and 3:

For question 2:
If a negative charge of double the magnitude of the first charge (20C) moves between the same two points in the electric field, we need to determine the change in electric potential (ΔV).

Since the electric field was not altered and the points remain the same, the electric potential difference will still be +10V. Therefore, the answer is 10V.

For question 3:
The change in electric potential energy (ΔPE) is given by the formula ΔPE = qΔV, where q represents the charge and ΔV represents the change in electric potential.

In this case, we have a negative charge of 20C (since it's double the magnitude of the first charge) and the change in electric potential (ΔV) of 10V.

Using the formula:

ΔPE = q * ΔV
ΔPE = -20C * 10V

Calculating:

ΔPE = -200J

Therefore, the change in electric potential energy based on question 2 is -200J.

I hope this clears up any confusion. Let me know if there's anything else I can help you with!