Electric Potential and Electric Field?

Question

Rank the arrangements in the picture according to the magnitude of the electric field at the center of the curvature, largest first.

http://img62.imageshack.us/img62/7821/92127235fc2.png

The instructor says that all of them have the same electric potential because they all have the same distance R and the same charge Q. But then, for electric field, the magnitude would be similar, just with an R^2 on bottom instead of just R. That would mean that all four are the same for electric field as well, right?.

But it looks to me that if the charge was evenly distributed the electric field on 4 should be 0, since it is pushing evenly on P from all directions and everything would cancel. Is this because E is a vector and therefore depends on direction as well as magnitude, while V only depends on magnitude? Or am I just not thinking straight and am totally off?

So how do you rank them?

Answer 1

indeed the electric field is zero for 4, because of the symmetry.
It is largest in 2, and so the ranking from top to bottom is:
2, 3,1, 4.

The electric potential is only defined as a the difference of potential energy per unit charge, so I assume that the other point is at infinity. In this case the addition is a scalar and all potentials are the same.

cheers,

Answer 2

Your thinking in second part of your question is correct.

If the charge was evenly distributed the electric field on 4 should be 0, since it is pushing evenly on P from all directions and everything would cancel. This is because E is a vector and therefore depends on direction as well as magnitude, while V only depends on magnitude.
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Draw a coordinate axes with the origin at the center of the circle O in figure 4.

There are 4 arcs one in each quadrant. The charges in each arc are Q/4. Assume that the charges are concentrated at the center of each arc. The magnitude of force at O due to each arc is K Q / [4 R^2]. As you thought, they cancel each other. Force is ZERO.

Considering the figure 1, there are two arcs alone. The charges in each arc are Q/2. As before assume that they are concentrated in their respective center of arc.
The magnitude of the force at O is K Q/ [2 R^2]. There are now two forces at right angles to each other. Their resultant is 2 cos 45 * K Q / 2R^2] =0.70 K Q / R^2, along the positive x direction.
This is more than the value in figure 4.

Considering the figure 3, there is only one arc. We can approximate it to 1/ 4 the of the circle. Let the x axis pass through the center of this arc. Each half of this arc has charge of Q/2. As before assume that the charges are concentrated at the center of the arc. There are two forces at O each of magnitude K Q/ [2 R^2] and inclined at angle of 45 degree to each other. Their resultant is 2* cos 22.5 * K Q/ [2 R^2] = 0.92 K Q/ R^2
This is more than the value in figure 1.

Considering the figure 1, the charges are concentrated at a point and the magnitude is the maximum = K Q /R^2.

The force is increasing gradually from zero to the maximum as the length of arc is reduced.

Rank in the ascending order is 4, 1, 3, and 2.

Answer 3

First, note the electric field is the -gradient (derivative) of the potential.

This means the electric field depends not on the value of the potential, but how that potential changes with respect to displacements in space.

So, if you integrate the field to find the potential on 4) you find that it varies with r, the distance from the center of the ring/sphere until you get to the surface (where the potential is 50kV relative to infinity, say). Once you are inside, then you are at a CONSTANT value of potential, e.g. 50kV. Since the potential is constant inside this region, the electric field is ZERO.

Thus knowing the value of the potential at a specific point is useless in finding the field. You kneed to know how that potential varies across SPACE.

Tachyon's rankings are correct.

Answer 4

I don't really know what I am talking about. It has been a long time since my physics. I'm going to take a shot.

Because we are talking about a DC charge it will be even. The greater "area"? of the circle would only amount of charge and not the charge itself (volts). Like a battery, it will remain at a specific voltage but changes the amount of charge with size.

Answer 5

it would be the same.