Finding energy in a 4 object charge system

[Anthony Santelices]

Homework Statement

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

Homework Equations

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

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.

 

[BvU]

Hello Anthony,

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

 

[ehild]

This is the picture :

Homework Statement

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

Homework Equations

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

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.

Have you left out the pair q1-q3?

 

[Anthony Santelices]

This is the picture :

View attachment 98121

Have you left out the pair q1-q3?

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.

Hello Anthony,

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

I don't have access to my work atm as I am not home. I will make sure to post it as soon I get the chance!

 

[Anthony Santelices]

Hello Anthony,

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

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

 

[ehild]

You ignored all the negative signs.

 

[Anthony Santelices]

You ignored all the negative signs.

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.

 

[ehild]

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