Work done by an external force to move a charged particle

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Homework Help Overview

The problem involves a charged particle moving in a circular path under the influence of another charged particle. The original poster seeks to derive an expression for the work done by an external force to increase the radius of the circular motion from r1 to r2, while considering the forces acting on the particle.

Discussion Character

  • Conceptual clarification, Assumption checking

Approaches and Questions Raised

  • The original poster attempts to relate the centripetal force and Coulomb force, noting a discrepancy in their calculations that results in a value three times that of a reference. Some participants question the signs used in the equations and the interpretation of forces involved in circular motion.

Discussion Status

Participants are exploring the relationship between centripetal and Coulomb forces, with some guidance provided on the nature of these forces. There is a recognition of the need to clarify the role of centripetal force in the context of the problem, but no explicit consensus has been reached.

Contextual Notes

There are indications of confusion regarding the direction of forces and their implications for the motion of the charged particle. The original poster's calculations appear to be influenced by assumptions about these forces.

maCrobo
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Homework Statement



A particle of (positive) charge Q is assumed to have a fixed position at P. A second particle of mass m and (negative) charge -q moves at constant speed in a circle of radius r1, centered at P. Derive an expression for the work W that must be done by an external agent on the second particle in order to increase the radius of the circle of motion, centered at P, to r2.

Homework Equations



W=∆U+∆K where U is the potential energy and K the kinetic energy.
K=\frac{1}{2}mv^{2}
U=\frac{Q(-q)}{4πε_{0}}\frac{1}{r}
F_{coulomb}=\frac{1}{4πε_{0}} \frac{Q(-q)}{r^{2}}
F_{centrifugal}=m\frac{v^{2}}{r}

The Attempt at a Solution


From the fact that F_{coulomb}=F_{centrifugal} I know that v changes as r changes. So I get the two values of v at r1 and r2.

Then I simply wrote W=∆K+∆U explicitly but my result is 3 times the book one. What may be wrong?
 
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Check your signs. It is the centripetal force which is equal to the Coulomb force, both pointing inward, towards the centre.

ehild
 
ehild said:
Check your signs. It is the centripetal force which is equal to the Coulomb force, both pointing inward, towards the centre.

ehild

Thanks for the answer, but there's something I can't get from your explanation.
If both forces pointed towards the centre, there would be a resultant inward force on the particle that make it accelerate along the radial direction, while it is actually still with respect to that direction.
 
The centripetal force does not "act". Performing circular motion, the particle accelerates towards the centre, this is the centripetal acceleration. The force needed to it is the centripetal force. The attractive force between the charges, the Coulomb force supplies the force needed for the circular motion. So you need to write that the centripetal force needed=Coulomb force (acting)

\frac{mv^2}{r}=k\frac{qQ}{r^2}

The centripetal force does not "act". It is not one force among other forces. It is the result of the real forces, which is needed to move a particle of mass m and speed v along a circle of radius r.

ehild
 
Last edited:
ehild said:
The centripetal force ... ... is the result of the real forces, which is needed to move a particle of mass m and speed v along a circle of radius r.

ehild

Sure! That's the point. Thanks a lot!
Then...
SOLVED!
 
Congratulation!:smile:

ehild
 

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