# Homework Help: Potential Difference in particle

1. Jul 11, 2008

### kdrobey

1. The problem statement, all variables and given/known data

A particle with a charge of -1.5 µC and a mass of 3.0 x 10-6 kg is released from rest at point A and accelerates toward point B, arriving there with a speed of 50 m/s.

(a) What is the potential difference VB - VA between A and B?

2. Relevant equations

v=(square root of) -2Qo(Va-Vb)/m
3. The attempt at a solution
everything else is given in the equation, v=50m/s, Q=-1.5 x 10^-6 C, m=3 x 10^-6, but i cannot figure out how to get Va-Vb by itself. I tried squaring both sides which gave me Va2=-4Qo2(Va-Vb)2/m, but from there i'm stuck

2. Jul 11, 2008

### kreil

Isn't it just algebra?

$$v= \sqrt{ \frac{-2Q_0 (V_a - V_b)}{m}}$$ (square both sides)

$$v^2 = \frac{-2Q_0 (V_A - V_B)}{m}$$(multiply/divide to move things around)

$$-\frac{ v^2 m}{2 Q_0}= V_A-V_B$$

3. Jul 11, 2008

### Redbelly98

Staff Emeritus
You might try thinking about conservation of energy (that is where the equation you give comes from anyway):

(Change in kinetic energy) = - (Change in electric potential energy)

4. Jul 11, 2008

### kdrobey

oh ok, i'll try that

5. Jul 12, 2008

### kdrobey

i plugged in the numbers and got -3 x 10^-6, but that was not right.

6. Jul 12, 2008

### Redbelly98

Staff Emeritus
I plugged in the numbers (Kreil's last equation in msg #2) and got something else. Try again, be careful about properly entering the numbers into your calculator.

Hint: the answer is a lot larger than 1V.

7. Jul 13, 2008

### kdrobey

ohh, a mental error on my part. got it, thanks!