# Electromagnetic Questions

Electromagnetic Questions :)

Hello everybody.

I have some questions that puzzled me in Electromagnetics. I tried to answer them all , some my answers make sene some not. I hope someone hear can help .

Q)If a material has a large permittivity, with it also follow that it will have a large conductivity ? Why or Why not ?

A)If the permittivity constant is larger since J=sigma.E so sigma=J/E so, If E is large then sigma will be large also.

I think this is wrong too.

Q)Which takes more energy to construct: a solid, uniform sphere of charge or a hollow uniform spherical shell with the same diamter and charge ? Why ? (think in terms of Energy density)

A) In terms of density a hollow uniformly sphere of charge will take more energy to construct. I
think becasue it has longer charge density than the solid sphere. the charge in the solid sphere is spread out within the solid spheere (more space for the charge) wich will be less charge density, the charge will be more compacted and highly dense in hollow sphere, becasue of less room for charge to (spread out)

My answer is wrong, I think I need to think in terms of Energy Density=1/2 DE , becasue my professor said its not the hollow.

Q) You have been placed in a room with a weightless point charge ans sensitive measuring devices. How could you use this charge to determine E and B fields at an arbitrary point in the room ?

A)With a test point charge I Wouild begin with Lorentz Force Law which states: F=Q(E+u x B)

Determine E: Since E=Fe/Q I would measure the force on Q(the test charge) This measurement would yield E.
Determine B: Since Fm=Qu x B I Would be looking at Q still, but this time I would more it at a velocity (u) and measure the force that is acting upon test charge (Q). B will be yielded operationally by undoing the cross product(how?????) . What velocity should I choose ?

## Answers and Replies

berkeman
Mentor
-1- Permittivity is usually largest for insulators. You get higher permittivity by having constrained electric dipoles in the material. Water has a high permittivity, and variable conductivity depending on impurities, however.

-2- My initial guess is that they are equal. The surface of each sphere is an equipotential surface, independent of whether the sphere is hollow or not.

-3- This is a trick question. There is no such thing as a "weightless point charge". Without mass, what would a force result in?

Thanks very much for helping me.

For the 2nd question, My professor stated that I need to figure to think of terms of (Energy Density = 0.5*D field . E field) to back it up mathematically.

For the 3rd question, I just wanted to know how how to undo the cross product. What velocity should I choose ?

Thanks very much :)

berkeman
Mentor
-2- Not sure I understand what you mean by energy density in this context. Do you mean the total amount of energy required to bring the net charge from infinity to the surface of the sphere(s)? Still should be the same in both cases, IMO.

-3- Let's assume the problem wording has been changed to "an electron" instead of a "weightless point charge". You need to have a mass to get an acceleration from the E and B fields. So you are mostly correct with the E-field answer, but you will measure the electron's acceleration, not the force on the point charge. I don't know of a practical way to measure the force directly. As for the B-field measurement, you don't have to "undo" the cross product. Let's say you get the electron moving somehow so it has a velocity v. Which direction is the resulting force from the B field going to act on the electron? The cross product shows you the vector situation. Let's say there is only a B field and no E field, and the electron has an initial velocity v in the x-direction. What is its motion going to look like? (Hint -- it will be a constant closed path motion....) And then what will be the general motion with both an E field and a B field at the same time?

Thanks very much. I understood measuring the B Field

About Electric Energy Density, I talked with my professor and he didnt give me clear full answer. He said it takes more energy to construct sold sphere of charge. Becasue we are doing work to push charges to the origin. Less energy for the shell (maximizing the distance from the charge on the surface to the origin).

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If a material has a large permittivity, with it also follow that it will have a large conductivity ? Why or Why not ?

I think the higher the value of the permittivity, the less conductance
How I can justify that by physical insight means in general , if my above statement is true?

Thanks

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berkeman
Mentor
See my statement -1- up in post #2. Look into the root of what it takes to make a good dielectric with high permittivity. And what quality do the charges have to have in a material that has a high conductivity?

To make good dielectric it should be a poor conductor of electricity, but an efficient supporter of electrostatic fields and metals have high conductivity casue metals possess charges that are free to move about the material as conduction currents.

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berkeman
Mentor
Correct. So if the charges are real free to move, it wouldn't seem to make much of a dielectric. What do you need for a good dielectric?

One caveat -- remember what I said about water. It's worth it for you to do a little reading about the dielectric constant of water, and how it might vary with conductivity due to impurities. I don't have any great links for it, but a quick google search should get you some stuff to read. Good luck!

would you show me how to find the capacitance per meter for this following geometry. My prof mentioned it but didnt talk much about it, he insisted that we do some reading on our own

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In the figure,. The corner surface along the x and y axis is held at potential of 0 Volts, and the hyperbolic surface y=1/x is held at potential of +4 Volts. The surfaces extend from –infinity < z < +infinity (out the paper) and extend towards infinity in the x and y directions, but the positive and negative surfaces never meet. The dielectric between the plates has a relative permittivity of €r= 4

a- Prove that the potential function between the conducting surfaces is V(r)= 4xy Volts where x and y in meters
b- Find the Voltage Vab between the points Pa(0.2,1,1) and Pb(1,0.5,2)
c- Find the surface charge density at the point (0,1,0)

#### Attachments

nrqed
Science Advisor
Homework Helper
Gold Member
berkeman said:
-1-

-3- This is a trick question. There is no such thing as a "weightless point charge". Without mass, what would a force result in?

Why??
One simply has to step in a spaceship very far from any star/planet. Ok, the spaceship would still create a gravitational force on the charge but it's tiny...and one could imagine being in a a spherical spaceship, near the center so that the weight is truly near zero.
weren't you rather thinking about "massless charge" ?

I think it makes sense to talk about weightless charges here since the presence of a gravtitational force would make it impossible to disentangle the effect of the weight from the effect of an electric field (without resorting to using other objects). So I don't think it was a trick question.

My suggestion would be : first, place the charge at a point and release it from rest. Does it start to move? If yes, there is an electric field in the direction of the motion at that point (if q>0). Notice that if there was a gravitational force, I would be stuck.

Next, set the chare in motion in the direction of the E field. Does it move along a straight line? Or does it start to curve?

In the first case you cannot conclude there is no B field, it could be parallel to the E field and thus cause no force for motion in that direction. You then have to repeat the experiment by giving the charge a velocity in a different direction than the E field and see if it has an acceleration determined only by the E field or if there is an extra acceleration (which will come from the B field)

Patrick

nrqed
Science Advisor
Homework Helper
Gold Member
DeltaX said:
In the figure,. The corner surface along the x and y axis is held at potential of 0 Volts, and the hyperbolic surface y=1/x is held at potential of +4 Volts. The surfaces extend from –infinity < z < +infinity (out the paper) and extend towards infinity in the x and y directions, but the positive and negative surfaces never meet. The dielectric between the plates has a relative permittivity of €r= 4

a- Prove that the potential function between the conducting surfaces is V(r)= 4xy Volts where x and y in meters
You are looking for a function V(x,y,z).
Now, this function is independent of z, by symmetry, so you really need a function V(x,y). In addition, from the graph, when y =1/x, V must be a constant equal to 4.

So V(x,y=1/x) =4.

Clearly, the function must be a function of the *product* xy, so you actually have V(x,y) = V(xy).

However, for infinite plates, the E fields are constant, so if we calculate $E_x = - {\partial V \over \partial x}$ let's say, we better get a constant. So the function must be linear in x. For the same reason it must be linear in y.

So that leaves us with V(x,y,z) = C_1 xy + C_2.

Imposing that V =0 at x=0 or y=0 gives C_2 =0. Imposing that V =4 at y=1/x gives C_1 = 4.

b- Find the Voltage Vab between the points Pa(0.2,1,1) and Pb(1,0.5,2)
Just plug in the coordinates in V(x,y,z) and subtract.
c- Find the surface charge density at the point (0,1,0)
The x and y components of the E fields are given by the partial derivatives I gave above. Also, you must have seen the equation relating the E field near the surface of an infinite plane to the surface charge density $\sigma$, right? You set the two equations for E_x equal and it will give you sigma of the vertical plate (which will be a function of y). Same for the horizontal plate.

Patrick

Thanks for the help. But I always been thinking we assumed its C1.XY We know there has to be multiplication but why in this form, it can be X/Y , Y^-1 . X ,.........

I was lookin at this problem and had some problem, would you guys please help ?

I Just worked it tru. Never mind :)

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