Electric Potential Homework: Answers to 1 & 2

AI Thread Summary
To calculate the work done in bringing a charge from infinity to a point in an electric field, the potential at that point is used directly, as it represents the work per unit charge. For the first problem, the potential at 4m is found to be 4495 V, and the work done can be calculated by multiplying this potential by the charge being moved. In the second problem, the final speed of protons released from a 5 MV potential is calculated to be 2.64 x 10^7 m/s. To find the accelerating electric field, the change in potential can be divided by the distance over which it occurs. Understanding these principles allows for the correct application of electric potential and field equations.
nns91
Messages
301
Reaction score
1

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.
 
Physics news on Phys.org
(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
 
I multiplied the values first without the error limit. Got 19.38. rounded it off to 2 significant figures since the given data has 2 significant figures. So = 19. For error I used the above formula. It comes out about 1.48. Now my question is. Should I write the answer as 19±1.5 (rounding 1.48 to 2 significant figures) OR should I write it as 19±1. So in short, should the error have same number of significant figures as the mean value or should it have the same number of decimal places as...
Thread 'A cylinder connected to a hanging mass'
Let's declare that for the cylinder, mass = M = 10 kg Radius = R = 4 m For the wall and the floor, Friction coeff = ##\mu## = 0.5 For the hanging mass, mass = m = 11 kg First, we divide the force according to their respective plane (x and y thing, correct me if I'm wrong) and according to which, cylinder or the hanging mass, they're working on. Force on the hanging mass $$mg - T = ma$$ Force(Cylinder) on y $$N_f + f_w - Mg = 0$$ Force(Cylinder) on x $$T + f_f - N_w = Ma$$ There's also...
Back
Top