Potential Energy of Point Charges

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Sneakatone
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Homework Statement


A point charge q2 = -4.5 μC is fixed at the origin of a co-ordinate system as shown. Another point charge q1 = 3.6 μC is is initially located at point P, a distance d1 = 7.3 cm from the origin along the x-axis



2) The charge q2 is now replaced by two charges q3 and q4 which each have a magnitude of -2.25 μC, half of that of q2. The charges are located a distance a = 1.7 cm from the oringin along the y-axis as shown. What is ΔPE, the change in potential energy now if charge q1 is moved from point P to point R?

Homework Equations


U= (kqQ)/r


The Attempt at a Solution


I don't know how to do the equations because 3 points confuse me.
but this is what I have.

(9E9*-2.25E-6*3.6E-6)(1/0.0749-1/0.073)=0.025

I got the 0.0749 from the hypotenuse and converted to meters.
 

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I take it you gave some answer in the 1) part and now continue with the 2) part of the exercise ?
Like 1) asking for the change in potential energy when q1 goes from d1 to d2 ? Did they give a value for d2 ?
And the answer number being your number, you didn't bother to show us what you did there ?
If I am wrong, correct me. If I am right, help me help you by telling what you did there.

Anyway, the problem formulation does not refer to the splitting up of q2 but instead it considers q3 and q4 in a fixed position and asks about moving q1 from d1 to d2. Your attempt at solution does contain something with d1 which is known, but I see nothing coming from d2 ?
The 1/(hypothenusa from d1) looks reasonable. That would be for one of the chaarges on the y axis. But there are wo of them!
The 1/d1 I don't see being very applicable any more.

By the time you have an answer, you want to compare it with the 1) answer. Check the ratio and explain wherer it comes from -- if my guesses on the 1) part weren't too far off.
 
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I should have posted number one:
1) What is ΔPE, the change in potenial energy of charge q1 when it is moved from point P to point R, located a distance d2 = 2.9 cm from the origin along the x-axis as shown?

The answer I got was -3.03 J

but I am still stuck
 
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0.073 is d1. from problem 1 d2 is 0.029 m. am I suppose to replace d1 with d2?
 
haruspex said:
Yes, but in problem 2 there never are two charges at distance d1. All that moves is q1, from P to R.
Yes. That was what I would want Sneak to conclude by him/herself. I think he/she can.
 
so that means the change in distance would be the same as the hypotenuse (0.073)?
 
BvU said:
Yes. That was what I would want Sneak to conclude by him/herself. I think he/she can.
It seemed to me Sneakatone had misread the question.
Sneakatone, in question 2, if we label the origin as O and the points where q3 and q4 are in the picture as A and B, where do you think the charges are (a) before the movement and (b) after the movement? Using the labels on the points, how far is q1 from the other charges in those two positions?
 
this is my drawing
 

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Sneakatone said:
this is my drawing
No doubt that diagram means something to you, but with no labels relating it to the OP it means nothing to anyone else.
In problem 2, there are always charges q3 and q4 at locations (0, -a) and (0, +a). At the start, q1 is at (d1, 0). At the end, q1 is at (d2, 0). How far was q1 from q3 before the move? How far is it from q3 after the move?
 
ok I see now its moving to point r to p.
so my set up is now (9E9*-2.25E-6*3.6E-6)(1/0.0749-1/0.0336).
0.0336 is obtained from the hypotenuse of d2 .
And I multiplied the whole equation by 2 because there are 2 of them there.
 
Sneakatone said:
ok I see now its moving to point r to p.
so my set up is now (9E9*-2.25E-6*3.6E-6)(1/0.0749-1/0.0336).
0.0336 is obtained from the hypotenuse of d2 .
And I multiplied the whole equation by 2 because there are 2 of them there.
Yes. I haven't checked whether you have the sign right. What sign do you expect the answer to have?
 
the answer I got is -2.392 which the sign is negative.
comparing to #1 the raito is1.266
im guessing the ratio came from the absolute value of q2/q1
 
Sneakatone said:
the answer I got is -2.392 which the sign is negative.
comparing to #1 the raito is1.266
im guessing the ratio came from the absolute value of q2/q1
No, it must have come from some difference in the two set-ups. The difference is the distribution of the negative charge. I don't think BvU is concerned about the precise value of the ratio, merely which is the larger (correct me if I'm wrong BvU). So the question is, why is the PE change less in the second set-up?