# Homework Help: Electric Potential and Kinetic Energy

1. Jul 25, 2015

### TAMUwbEE

1. The problem statement, all variables and given/known data
A charged particle (q = -8.0 mC), which moves in a region where the only force acting on the particle is an electric force, is released from rest at point A. At point B the kinetic energy of the particle is equal to 4.8 J. What is the electric potential difference VB-VA?

2. Relevant equations
EPE = Delta V (volts) = Delta U (potential energy) / q0

3. The attempt at a solution
Using this equation, I came up with the solution of -.6 kV, however, the answer is positive. Why is this?

2. Jul 25, 2015

### Qwertywerty

Which point do you think is at a higher potential ?

3. Jul 25, 2015

### TAMUwbEE

Ahhh. Point VB has higher potential energy, therefore, in the equation VB - VA, it is positive. Thank you for your response.

I also have another question.

A particle (charge = +2.0 mC) moving in a region where only electric forces act on it has a kinetic energy of 5.0 J at point A. The particle subsequently passes through point B which has an electric potential of +1.5 kV relative to point A. Determine the kinetic energy of the particle as it moves through point B.

I'll be honest with you, I do not know where to start with this problem. I know my equations however, I cannot picture the concept of potential energy caused by an electric field in relative to points A and B. If someone could clarify I would greatly appreciate it. Thank you.

4. Jul 25, 2015

### Qwertywerty

Simply , q*ΔV = ΔKE .

5. Jul 25, 2015

### TAMUwbEE

Thanks again Qwerty. I must have missed out on that equation when I was copying my notes.

6. Jul 25, 2015

### TAMUwbEE

1. The problem statement, all variables and given/known data
Identical point charges (+50 μC) are placed at the corners of a square with sides of 2.0-m length. How much external energy is required to bring a fifth identical charge from infinity to the geometric center of the square?

2. Relevant equations
Electric Potential for Multiple Charges

ΔU = K∑(Qi / Ri)
Therefore,
K(50x10^-6 C) / 2cos(45°) yields 3.17 x 10^5 V.

3. The attempt at a solution
Each point charge by itself has an electric potential of 3.17 x 10^5 V (V is the correct unit, right?).

However, I'm sure I'm doing this wrong because my answer choices are in a completely different magnitude (10^1).

Thanks for the help.

7. Jul 25, 2015

### Qwertywerty

ΔU = K∑(Qi*Q/Ri) .

Also , I think you haven't inputted Ri correctly either .

8. Jul 25, 2015

### TAMUwbEE

What should the equation be then for all 4 charges?

K(Q1 + Q2 + Q3 + Q4) / 2cos(45)?

Also, what would Ri be then?

9. Jul 25, 2015

### Qwertywerty

Ri equals distance between any charge and the new charge ( Use diagram ) .

KQ(Q1 + Q2 + .. ) .

10. Jul 25, 2015

### haruspex

Are you sure? sqrt(2) looks right to me.
TAMUwbEE, I'm not sure whether you have understood Qwertywerty's other point. You calculated 3.17 x 10^5 V (as the potential due to one existing charge). You are asked for an energy, not a potential. That explains the orders of magnitude discrepancy.

11. Jul 25, 2015

### Qwertywerty

Then that's my mistake .