# work to move a charge in an electric field

Discussion in 'Physics' started by davebee, Nov 14, 2012.

1. ### davebee Thread Starter Well-Known Member

Oct 22, 2008
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1. suppose you have a charged conducting sphere with a hollow center and a hole in its side leading to the center, like the metal piece at the top of a van de Graaf generator. The electric field strength at the surface decreases to zero as the surface is followed to the interior of the hollow.

Does it take the same amount of work to move a test charge from a great distance away to a point on the surface at the center of the hollow as to move the charge to a point on the outer surface?

2. similar to #1 except this time you have a charged conducting sphere with a conducting needle with a sharp tip sticking out from its side. The electric field strength becomes very intense near the pointed tip of the needle.

Same question as #1 - does it take the same amount of work to move a test charge from a great distance away to the tip of the needle as to move the charge to the smoothly rounded surface?

In both cases, I think the answer is yes, but what physics principle answers the questions?

I'm assuming that even though the field lines are visualized as closer together near the needle tip and further apart (to the point of not existing) near the hollow inside, each line is equivalent to any other line in terms of the work that must be done in moving a test charge from a great distance away, along each line to the surface of the object. Does that sound correct?

2. ### russ_hensel Distinguished Member

Jan 11, 2009
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There is no field inside the sphere.

This all depends on the voltage of the sphere, look at the definition of electric potential ( measured in volts, often called voltage )

3. ### davebee Thread Starter Well-Known Member

Oct 22, 2008
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Never mind, I think I was being misled by statements like this, which can be found in many introductory physics lessons:

"The surface of a conductor is an equipotential surface, i.e. the electric potential is the same at every point of the surface."

I think it really means that the vector sum of the HORIZONTAL components of the electric field is zero for every location on the surface; otherwise charge would move until the HORIZONTAL component of the field is zero.

It's saying nothing about the electric field strength ABOVE the surface, where for the tip of a needle point, the perpendicular strength of the electric field may be much stronger than the field on the smooth part of the object, meaning it will take more work to bring a test charge to the tip, meaning the tip is at a higher voltage than the smooth surface of the object.

I suspect the case of the charge moving into the hollow center is even more complicated. It will take work to move the charge up to the entrance of the hollow. At some point, there will be no force on the charge. Then the charge will start inducing opposite charges on the inner surface of the object, which will attract the test charge to the inner surface, where it will neutralize the induced charges and leave the hollow center neutral again.

4. ### WBahn Moderator

Mar 31, 2012
23,587
7,215
So consider this:

You have a conducting sphere with a needle sticking out. If the voltage (remember, voltage is a scalar, not a vector) at the tip of the sphere is different from the voltage at the opposite side of the sphere where it is smoothest, why won't there be a current flowing from the needle tip toward the other side of the sphere? If the answer is that there would be, then what does that say about the voltage at all points on the exterior surface of this conductor, including the needle tip?

Now, it is certainly true that the tangential component of the electric field must be zero at the surface of a conductor in an electrostatic case and that there is no requirement that the normal component be uniform. But electric field and voltage are not the same thing.

5. ### davebee Thread Starter Well-Known Member

Oct 22, 2008
539
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Voltage is a scalar, but it derives its meaning from a vector, the electric field.

A location on the needle tip (or anywhere on the object) has a voltage due to the normal component of the electric field. But since the tangential component of the electric field has reached a balance such that the net field is zero at each charge, then there is no force to move any charge in any direction along the surface (once the charges have reached their static configuration).

So there will be no current along the surface, even though it can be said that points at different locations on the surface have different voltages.

6. ### WBahn Moderator

Mar 31, 2012
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7,215
So you are saying that, in a static situtation, you can have a conducting path between two different voltages and not have any current flow that results? Does that really make sense to you?

Look at the definition of voltage. It is the line integral of the component of the electric field along that line. If your line integral is along the surface of the conductor, then the component of the electric field is along that line is the trangential component of the electric field at the surface of the conductor. But you yourself just said that the charges have arranged themselves so as to make this component zero. Hence, there is no voltage difference between any two points in the surface of a conductor under electrostatic conditions.

7. ### davebee Thread Starter Well-Known Member

Oct 22, 2008
539
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No, it doesn't, but what else can it be?

The electric field normal to the surface is stronger near the needle tip than it is near the smooth body. Voltage is the integral of the field over a path.

So if you integrate over the field from say Earth to the needletip, then integrate over the field from Earth to the smooth surface of the object, I don't see why you won't get different values.

The only way you wouldn't would be if the field further from the needletip weakened just enough to balance the extra strong field near the tip, to result in the integral having the same value as over the path from Earth to the smooth part, and I don't see any reason for any additional weakening to take place in the path between the needletip and Earth as opposed to the path from the smooth body to Earth.

Which results in the odd case that there is a different voltage from Earth to the two locations on the object, yet there is no voltage along the surface between the two locations on the object.

Am I wrong? And if you think so, can you explain the physics for why?

8. ### WBahn Moderator

Mar 31, 2012
23,587
7,215
You actually have the answer explained fairly well. You have an extra strong field near the sharp tip, but the field drops off very quickly compared to the rate at which it drops off for the flatter portions. As a result, the field further way straight out from the sharp point is actually lower than the field straight out from the flatter areas. But the voltage will still be lower (i.e., have dropped more) out from the sharp point than from the flatter region, it just won't be dropping as much. It's like having two paths from the top of a mountain down to a campsite. One path might be a straight line while another path might be a concave curve. On the curved path you start off with a steep descent (high electric field) and quickly end up at a much lower elevation (voltage) than the straight path does. But, after a while, your rate of descent is the same as that of the straight path, so you have equal electric fields at that point, but you are at a significantly lower elevation (lower voltage). However, beyond this point both paths are still descending (dropping in voltage) but the straight path is now dropping quicker (higher electric field) and, at the campsite (at infinity) has exactly caught up.

Why? Because the campsite is at a single elevation (voltage at infinity is zero) and the path integral of the rate of descent (of the electric field) in the direction of the path along any path between the mountaintop and the campsite (the surface and infinity) has to be the same.

Far from the object, the field looks like that from a point charge equal to the total charge on the sphere. The thing to keep in mind is that while the charge density is higher near parts with small radii of curvature, the smaller charge densities are over larger areas and hence the total charge in those areas is larger.

So imagine a uniformly charged sphere that has spherical equipotential surfaces. As you add a tip to the sphere, the equipotential surfaces are distorted and pulled down toward the tip. Imagine being inside a rubber balloon and reaching up to the inner surface and pulling it straight toward you. You create an inward dimple that smooths out the further sideways you go and, when you are far enough away (on the balloon's surface) from the center of the dimple, the less distortion you notice. Now imagine lots of balloons, each inside the other. As you move outward to the next larger balloon, you still make a dimple, but you don't pull it in as hard. Eventually, you are barely creating a dimple at all and, at that point, you are in the "far field".

Another way to imagine it is to go the opposite direction: The balloon that is nearest the charged object is basically sucked down right against it since the surface of the object is an equipotential surface. The next balloon out isn't sucked down quite as hard and so it is trying to return to a sphere. The points of highest curvature are suckiing down the hardest so, in those regions, the balloon is pulled closer to the previous balloon. In the other regions, the balloon is able to be further away and closer to a sphere. As you move out, the balloons become less and less distorted. At any point, the electric field is inversely proportional to the distance between adjacent layers of balloon.

9. ### davebee Thread Starter Well-Known Member

Oct 22, 2008
539
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Ok, I'm starting to think you're right, but a different explanation occurred to me.

Imagine a uniform sphere, surrounded by balloons representing equipotential shells. Initially the balloons will also be spheres.

Now imagine a needlepoint rising from the surface, but not penetrating the balloons. The needlepoint will press the balloon layers upward, packing them against the next higher layers.

A little beyond the needlepoint, the balloon layers will show a slight outward bulge, but far from the point, the layers will remain spherical.

Integrating the path through the electric field to the needlepoint encounters a region of stronger field, but since the needlepoint is raised up from the surface, the integration to the needlepoint takes place over a shorter distance than the integration to the smooth surface.

This could account for the integration to the two locations giving the same result, but without requiring a region of lesser field beyond the needlepoint.

I could easily imagine that the two factors of stronger field but over a shorter distance will result in a total integral equal to that of the integral path to the smooth surface.

Does this make sense?

10. ### WBahn Moderator

Mar 31, 2012
23,587
7,215
I think you've got the idea. Whether the equipotentials are pulled closer or pushed out depend on the exact shape we are talking about and also what we use as our reference sphere to start with.