Work done by an external force to move a charged particle

In summary, the external agent must perform work on the second particle in order to increase the radius of the circle of motion, centered at P, to r2.
  • #1
maCrobo
51
1

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.
[itex]K=\frac{1}{2}mv^{2}[/itex]
[itex]U=\frac{Q(-q)}{4πε_{0}}\frac{1}{r}[/itex]
[itex]F_{coulomb}=\frac{1}{4πε_{0}} \frac{Q(-q)}{r^{2}}[/itex]
[itex]F_{centrifugal}=m\frac{v^{2}}{r}[/itex]

The Attempt at a Solution


From the fact that [itex]F_{coulomb}=F_{centrifugal}[/itex] 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?
 
Physics news on Phys.org
  • #2
Check your signs. It is the centripetal force which is equal to the Coulomb force, both pointing inward, towards the centre.

ehild
 
  • #3
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.
 
  • #4
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)

[tex]\frac{mv^2}{r}=k\frac{qQ}{r^2}[/tex]

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:
  • #5
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!
 
  • #6
Congratulation!:smile:

ehild
 

1. What is work done by an external force to move a charged particle?

Work done by an external force to move a charged particle is the amount of energy transferred to the particle by the force in order to move it from one point to another. This work is calculated by multiplying the magnitude of the force by the displacement of the particle in the direction of the force.

2. How does the work done by an external force affect the speed of a charged particle?

The work done by an external force on a charged particle can change its speed by increasing or decreasing it. When the force is applied in the direction of motion, the particle's speed will increase. If the force is applied in the opposite direction of motion, the particle's speed will decrease.

3. Can the work done by an external force be negative?

Yes, the work done by an external force can be negative. This happens when the force is applied in the opposite direction of the displacement of the particle. In this case, the force is doing negative work, which means it is taking energy away from the particle and decreasing its speed.

4. How does the angle between the force and displacement affect the work done on a charged particle?

The angle between the force and displacement affects the amount of work done on a charged particle. When the force and displacement are in the same direction, the work done is maximum. If the force and displacement are perpendicular, the work done is zero. And if the force and displacement are in opposite directions, the work done is negative.

5. What is the relationship between work done by an external force and the change in kinetic energy of a charged particle?

The work done by an external force on a charged particle is equal to the change in its kinetic energy. This means that if the work done is positive, the kinetic energy of the particle will increase. If the work done is negative, the kinetic energy of the particle will decrease. And if the work done is zero, there will be no change in the kinetic energy of the particle.

Similar threads

  • Introductory Physics Homework Help
Replies
12
Views
206
  • Introductory Physics Homework Help
Replies
14
Views
639
  • Introductory Physics Homework Help
Replies
9
Views
926
  • Introductory Physics Homework Help
Replies
5
Views
742
  • Introductory Physics Homework Help
Replies
4
Views
815
  • Introductory Physics Homework Help
Replies
23
Views
350
  • Introductory Physics Homework Help
Replies
3
Views
801
Replies
22
Views
1K
  • Introductory Physics Homework Help
Replies
17
Views
404
  • Introductory Physics Homework Help
Replies
11
Views
1K
Back
Top