Question on finding work done on charge

In summary, the conversation discusses the approach to finding the amount of work required to bring a charge q3 from infinity to its current position in a system with two other charges, q1 and q2. The discussion revolves around the use of Coulomb's law and the integration of the y-component of the electric force. The solution shown uses the potential to calculate the potential energy of q3, which is found to be independent of the path taken. This is due to the fact that the electric field is static and the potential function is scalar and additive. Ultimately, the conversation concludes that the work required is the same regardless of the method used.
  • #1
RoboNerd
410
11

Homework Statement




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Homework Equations



coulomb's law

The Attempt at a Solution



Hi everyone. I understand their approach with the integration to find the amount of work that "a person" would have to do to bring the charge q3 from infinity to its current position.

I understand that the force that this person would have to do is the opposite of the coulombic forces between the q3 and q1 and 2.

My question: why did they not account for the y-components of the coulombic forces between q1 and 2 to a3? The x-components of the two forces would cancel, so the only coulombic forces acting on q3 would be the y-components.

Thus I did exactly what they did, but I added a sine term to account for the y-component and then re-wrote the sine in terms of the vertical distance and "s/2" with pythagorean theorem and then did the integration.

Could anyone please weight in on why their approach is legit and they did not do what I planned to do with the sine term?

Thank you very much in advance.[/B]
 
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  • #2
If you correctly do the integration of the y component of the electric force you get the same result.
The solution shown uses the potential to get the potential energy of q3, which is q3U. The potential of a point charge is kq/r where r is the distance from the charge. The potential function is scalar and additive, so the net potential at a point is the sum of the potentials from all charges.
 
  • #3
I think what is needed to be emphasized is that the work is independent of the path that the q3 follows as is brought from the infinite to the desired place, it depends only on the end points of the path (one point at infinite and one point at the third vertex of the triangle). This holds only if the field is static so that ##\nabla \times E=0##, ##E=\nabla\phi## where ##\phi## is the electrostatic potential (scalar).
 

FAQ: Question on finding work done on charge

How is work done on a charge calculated?

The work done on a charge is calculated by multiplying the magnitude of the charge by the potential difference it moves through. This can be expressed as W = qΔV, where W is the work done, q is the charge, and ΔV is the potential difference.

What is the unit of work done on a charge?

The unit of work done on a charge is joules (J). This is the same unit as for energy, as work done is a measure of energy transferred.

Can the work done on a charge be negative?

Yes, the work done on a charge can be negative. This occurs when the charge moves in the direction opposite to the direction of the electric field, resulting in a decrease in potential energy. In this case, the work done is considered as negative to indicate the reduction in energy.

How does the direction of the charge affect the work done?

The direction of the charge does not affect the calculation of work done. However, the sign of the work done will be positive if the charge moves in the same direction as the electric field, and negative if the charge moves in the opposite direction.

What are some real-life examples of work done on a charge?

Some real-life examples of work done on a charge include charging a battery, powering electronic devices, and electric motors converting electrical energy into mechanical energy. Work done on a charge is also involved in processes such as electrolysis and lightning strikes.

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