Question regarding electrical potential energy required to assemble a system

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  • #1
Parad0x88
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Homework Statement


Four charges, q1=+q, q2=-q, q3=+q and q4=*q are at the corners of a square with the side a=7.5 cm. (In such way that if one travels along the perimeter of the square, the charge signs are alternating.) If q=3.0 μC, what is the total energy required to assemble this system of charges? Comment on the sign of your answer.


Homework Equations


U = Ke ((q1q2/r12) + (q1q3/r13) + (q1q4/r14) + (q2q3/r23) and so forth)


The Attempt at a Solution


Correct me if I'm wrong, but I believe I simply have to sum up the energy between the points one by one and multiply by Coulomb's constant, and it will give me the total potential energy?
 

Answers and Replies

  • #2
Doc Al
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You got it.
 
  • #3
Parad0x88
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You got it.

Thanks!

My question is: Would I need to calculate individually q1q4/r14 and q4q1/r41, or since they are the same I only do it once?
 
  • #4
Doc Al
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My question is: Would I need to calculate individually q1q4/r14 and q4q1/r41, or since they are the same I only do it once?
Count the potential energy between each pair of charges only once.
 
  • #5
Parad0x88
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Ok, thank you!

And last but not least, if my energy is negative (I got -2,788,518 joules) that basically means that negative work is done when assembling the system to avoid that the charges will move toward each other?

And do you mind verifying my formula to make sure I didn't do anything wrong? I got an obscurely high number as an answer:

So like I had said: U = Ke ((q1q2/r12) + (q1q3/r13) + (q1q4/r14) + (q2q3/r23) + (q2q4/r24) + (q3q4/r34))

r12, r14, r23 and r34 are 0.075m
r13 and r24 are: √(0.0752 + 0.0752) = 0.106m

So I get:

U = Ke ((3 * -3/.075) + (3 * 3/.106) + (3 * -3/.075) + (-3 * 3/0.075) + (-3 * -3/.106) + (3 * -3/.075))

U = Ke ((4X (-9/.075)) + (2X (9/.106))

U = Ke ((4 X -120) + (2 X 84.91))

U = Ke(-310.18 μC2/m)

U = 8.99 X 109 Nm22/C2 X -310.18 μC2/m

U = -2,788,518.2 Nm, or joules
 
  • #6
Doc Al
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And last but not least, if my energy is negative (I got -2,788,518 joules) that basically means that negative work is done when assembling the system to avoid that the charges will move toward each other?
Right.

And do you mind verifying my formula to make sure I didn't do anything wrong? I got an obscurely high number as an answer:

So like I had said: U = Ke ((q1q2/r12) + (q1q3/r13) + (q1q4/r14) + (q2q3/r23) + (q2q4/r24) + (q3q4/r34))
Good.

r12, r14, r23 and r34 are 0.075m
r13 and r24 are: √(0.0752 + 0.0752) = 0.106m
Good.

I think you're failing to multiply the charges correctly and thus missing a factor of 10-6. For example:
q1*q2 = (3 x 10-6)*(-3 x 10-6) = -9 x 10-12
 
Last edited:
  • #7
Parad0x88
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Right.


Good.


Good.

I think you're failing to multiply the charges correctly and thus missing a factor of 10-6. For example:
q1*q2 = (3 x 10-6)*(-3 x 10-6) = 9 x 10-12

Ah you're right! Rookie mistake of wanting to save some time and not writing the units hahah

Thank you very much!
 

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