How Much Work is Needed to Bring Two Point Charges Together?

AI Thread Summary
To calculate the work needed to bring two point charges, Q1 = 3.3 µC and Q2 = 7.6 µC, from a large distance to a separation of 2.6 m, the potential energy change is calculated using the formula ΔPE = (k Q1 Q2) / r. The final potential energy is found to be 0.0868 J, while the initial potential energy is effectively zero at infinite separation. The work done is then determined to be positive, as it is equal to ΔPE, not negative as initially thought. The discussion clarifies that the external force must counteract the repulsive Coulomb force, resulting in positive work. The correct interpretation emphasizes that the work required to bring the charges together is indeed 0.0868 J.
rbraunberger
Messages
9
Reaction score
0

Homework Statement



Two point charges Q1 = 3.3 µC and Q2 = 7.6 µC are initially very far apart. They are then brought together, with a final separation of 2.6 m. How much work does it take to bring them together?

Homework Equations



\DeltaPE = ( k Q1 Q2 ) / r

Work = - \DeltaPE

k= 8.99 e 9

The Attempt at a Solution



PE final = k Q1 Q2 / r
= (8.99e9)(3.3e-6)(7.6e-6) / ( 2.6m)
= 0.0868 J

PE initial = k Q1 Q2 / r
with r initial being infinitely large PE initial is basically 0

so \DeltaPE = 0.0868 + 0
Work = - \DeltaPE = -0.0868What is the error in this? Thank you!
 
Last edited:
Physics news on Phys.org
Is the separation 2.6 m or 2.8 m? And do not forget the unit of work!ehild
 
It was 2.6m...I fixed it. I found that the answer was just 0.0868 not -0.0868. My teacher said the equation was Work = -\DeltaPE, but I found it some place else that shows it as just Work = \DeltaPE. Which is correct?
 
The work needed to bring the charges together was the question. It is some external force Fe (maybe your force) that does this work. The external force is opposite and at least equal in magnitude with the Coulomb force Fe=-Fc. As the charges repel each other, the external force must push them toward the centre, so the direction of the external force is the same as the displacement: the work is positive.

ehild
 
Kindly see the attached pdf. My attempt to solve it, is in it. I'm wondering if my solution is right. My idea is this: At any point of time, the ball may be assumed to be at an incline which is at an angle of θ(kindly see both the pics in the pdf file). The value of θ will continuously change and so will the value of friction. I'm not able to figure out, why my solution is wrong, if it is wrong .
Thread 'Voltmeter readings for this circuit with switches'
TL;DR Summary: I would like to know the voltmeter readings on the two resistors separately in the picture in the following cases , When one of the keys is closed When both of them are opened (Knowing that the battery has negligible internal resistance) My thoughts for the first case , one of them must be 12 volt while the other is 0 The second case we'll I think both voltmeter readings should be 12 volt since they are both parallel to the battery and they involve the key within what the...
Thread 'Trying to understand the logic behind adding vectors with an angle between them'
My initial calculation was to subtract V1 from V2 to show that from the perspective of the second aircraft the first one is -300km/h. So i checked with ChatGPT and it said I cant just subtract them because I have an angle between them. So I dont understand the reasoning of it. Like why should a velocity be dependent on an angle? I was thinking about how it would look like if the planes where parallel to each other, and then how it look like if one is turning away and I dont see it. Since...
Back
Top