Electric Potential Homework: Answers to 1 & 2

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SUMMARY

This discussion addresses two physics homework problems related to electric potential and work done by electric fields. For the first problem, the work required to bring a 3 microC charge from infinity to a distance of 4 meters from a fixed 2 microC charge is calculated using the formula V = kq/r, resulting in a potential of 4495 V. The work done is determined by multiplying the potential difference by the charge, eliminating the need for integration. The second problem involves protons from a Van de Graaff accelerator, where the final speed of 5-MeV protons is calculated as 2.64 x 107 m/s, and the electric field is derived from the potential change over a distance of 2 meters using E = -dV/dx.

PREREQUISITES
  • Understanding of electric potential (V = kq/r)
  • Knowledge of work-energy principles in electrostatics (U = qV)
  • Familiarity with electric fields and their relationship to potential (E = -dV/dx)
  • Basic calculus for integration and differentiation in physics contexts
NEXT STEPS
  • Study the concept of electric potential energy and its applications in electrostatics.
  • Learn about the relationship between electric fields and potential differences in uniform fields.
  • Explore the principles of particle acceleration in electric fields, particularly in Van de Graaff accelerators.
  • Investigate the integration of electric fields to calculate work done in various scenarios.
USEFUL FOR

Students studying electromagnetism, physics educators, and anyone preparing for exams involving electric potential and work in electric fields.

nns91
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Homework Statement


1. A positive charge of magnitude 2 microC is at the origin.
(b)How much work must be done by an outside agent to bring a 3 microC charge from infinity to r=4, assuming that the 2 microC charge is held fixed at the origin ?
(c) How much work must be done by an outside agent to bring the 2 microC charge from infinity origin if the 3 microC charge is first placed at r=4m and is then held fixed ?

2. Protons from a Van de Graaff accelerator are released from rest at a potential of 5 MV and travel through a vacuum to a region at zero potential.
(a). Finad the final speed of the 5-MeV protons.
(b). Find the accelerating electric field if the same potential change occurred uniformly over a distance of 2m


Homework Equations



V=kq/r
E= -dV/dx
U=qV

The Attempt at a Solution



1. So I found that the electric potential at 4m is 4495 V and at infinity V=0.

To find the work done, I have to integrate -Udx but I don't know what to put for the lower limit. Do I put infinity ?

Or How do I do part b and c ?

2. So I found the speed in part a using energy equation and got 2.64 x 10^7 m/s

How do I then do part b ? I know that E= -dV/dx but then I cannot solve this because I don't know dV.
 
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(It's been a long time since I last looked at this kind of question, so this post may be from moderately to outrageously wrong)First, the electric potential at point r is: V=\frac{k|Q|}{r^2} V

Assuming we're working on vaccum, k\approx9\cdot10^9 N\cdot m^2/C^2 and V\approx1125 V.

Potential energy is Ep=q\cdot V J, so Ep\approx3.375\cdot10^{-3} J. That's 1.b.

1.c follows the same calculations, only exchanging both charges. The potential energy is the same.

2.a and 2.b: Have no friggin' clue ^^'.
 
Last edited:
nns91 said:
1. So I found that the electric potential at 4m is 4495 V and at infinity V=0.

To find the work done, I have to integrate -Udx but I don't know what to put for the lower limit. Do I put infinity ?

The potential is work already, you do not need to integrate. Assuming zero potential at infinity, the potential at a point is the work done by the electric field on a unit positive charge when the charge moves from infinity to that point. So to get work from potential difference, you simply multiply it with the charge.
If you calculate work by integrating the electric field intensity, the lower limit is the place where you know the potential. It is infinity here.
For question b, remember that r in the formula for the potential is the distance from the fixed charge. The two charges are 4 m apart again, so r=4 m.

nns91 said:
2. So I found the speed in part a using energy equation and got 2.64 x 10^7 m/s

How do I then do part b ? I know that E= -dV/dx but then I cannot solve this because I don't know dV.

If the electric field is uniform over a distance D, dV/dx = change of potential / D.

ehild
 

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