# Homework Help: Finding energy in a 4 object charge system

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1. Mar 28, 2016

### Anthony Santelices

1. The problem statement, all variables and given/known data
An object with charge +2.0×10−5 C is moved from position C to position D in the figure (Figure 1) . q1 = q3 = +10.0×10−5 C and q2 = −20.0×10−5 C. All four charged objects are the system.

Here's a picture to the problem

2. Relevant equations
$$F = \frac {kq_1 q_2}{r^2}$$
$$U_e = \frac {kq_1 q_2}{r}$$
$$a^2+b^2=c^2$$

3. The attempt at a solution
I first attempted finding all the energy for the system by finding the initial potential electric energy for each pair of charged objects and added them together. I managed to find the distance from $q_2$ to $q_1$ and $q_3$ using Pythagorean theorem.$$c → q_1, c → q_2, c → q_3, q_2 → q_1, q_2 → q_3$$
Unfortunately after putting the added energies it came up as wrong.

2. Mar 28, 2016

### BvU

Hello Anthony,

I suspect something went wrong with your calculations, can you show them, please ?
Note that 'the system' consists of four charges!

3. Mar 28, 2016

### ehild

This is the picture :

Have you left out the pair q1-q3?

4. Mar 28, 2016

### Anthony Santelices

Ah yes! It appears I did. However, even without having it I was getting 126 J. It says that the answer is -105 J. So that tells me I messed up somewhere while calculating with all the $U_e$'s.
I dont have access to my work atm as I am not home. I will make sure to post it as soon I get the chance!

5. Mar 28, 2016

### Anthony Santelices

These are the numbers I got along with the added $q_1, q_3$
$$U_{c,q1} = \frac {(9*10^9)(+2.0*10^{-5}) (+10.0*10^{-5})} {2.0} = 9\ J$$
$$U_{c,q2} = \frac {(9*10^9)(+2.0*10^{-5}) (-20.0*10^{-5})} {2.0} = 18\ J$$
$$U_{c,q3} = \frac {(9*10^9)(+2.0*10^{-5}) (+10.0*10^{-5})} {2.0} = 9\ J$$
$$U_{q2,q1} = \frac {(9*10^9)(-20.0*10^{-5}) (+10.0*10^{-5})} {2\sqrt2} = 45\ J$$
$$U_{q2,q3} = \frac {(9*10^9)(-20.0*10^{-5}) (+10.0*10^{-5})} {2\sqrt2} = 45\ J$$
$$U_{q1,q3} = \frac {(9*10^9)(+10.0*10^{-5}) (+10.0*10^{-5})} {4.0} = 22.5\ J$$

6. Mar 29, 2016

### ehild

You ignored all the negative signs.

7. Mar 29, 2016

### Anthony Santelices

I've tried it with the negative signs, but I end up getting -68. The correct answer shown was -105, so I don't know exactly where I went wrong.

8. Mar 29, 2016

### ehild

You must include the signs of the charges,
And even the numerical values of U(q2,q1) and U(q2,q3) are wrong.