# Tag Info

4

In this case, the image method can be used to calculate the potential (and hence the electric field) in the region $z>0$, with a negative charge $-q$ located at $(0,0,-d)$, since the potential would be $V(x,y,0)=0$, in this case. But for points in the region $z<0$, the potential is given by the solution of Lapalace equation $\nabla^2 V=0$, with ...

4

I think you are reading a lot into what is a minor distinction. Strictly speaking I suppose the gravitational potential is the energy per unit mass, i.e. $m=1$ in your first equation, while the gravitational potential energy is the potential times the mass. In practice no-one I know has ever bothered to make the distinction because it's usually obvious what ...

3

I too was confused by this difference between gravity and electromagnetism. Hopefully the following clears things up. The gravitational potential a distance $r$ from a mass $M$ is $$\phi_g=-\frac{GM}{r},$$ the gravitational field is $${\bf g} = - \nabla \phi_g,$$ and the gravitational potential energy (of two masses $M$ and $m$ separated by a distance ...

2

What's probably happening here is the following: The fundamental or microscopic fields $\mathbf{E}$ and $\mathbf{B}$ are technically called the electric field strength and the magnetic induction, while $\mathbf{D}$ and $\mathbf{H}$, their macroscopic counterparts, are called the electric displacement and the magnetic field, a quite weird nomenclature, since ...

1

Well, the electric field $\vec E$ is different from the force field $\vec F$ a test charge will feel. That difference is exactly the charge of the test particle. That force field is given by the gradient of a function, too $$q \vec E = \vec F = - \frac{\mathrm d}{\mathrm d r} W$$ where I use the letter $W$ in order not to have confusing notation. The ...

1

Yes! Neutrons are electrically neutral, but they have a magnetic moment. You can accelerate a bar magnet with a magnetic field, so you can also accelerate a neutron with a magnetic field. For most beams, the change in energy is pretty negligible, but there's a major exception for ultra-cold neutrons (UCN), which just so happen to be my specialty. UCN have ...

1

Electric field is defined as the electric force per unit charge. The direction of the field is taken to be the direction of the force it would exert on a positive test charge. So, if there is force acting on a unit charge, then electric field does exist. It is the way by which we can prove the existence of electric field (as per definition demands). I don't ...

1

No, there are several mistakes in your derivation, although you miraculously end up with the right expression. The term $\sin \theta$ comes from taking the horizontal (X) component of the electric field - not from the expression you used for $dq$. The diagram I envisage for your problem is this: So I would say that $$dq = \lambda dy$$ and then dE = ...

1

Your method for evaluating the electric field assumes it is appropriate to model it as spatially constant within the wire since you're basically taking a spatial average. You'll have to decide wether or not this is accurate.

1

Therefore, the energy released by allowing the distance between the plates to slowly decrease to zero is U=ϕQ This isn't correct. While it is true that the electric field magnitude between the plates is $E = \frac{\sigma}{\epsilon_0}$, this is the sum of the two electric fields from both plates. But the force on the charge on one plate is due to the ...

Only top voted, non community-wiki answers of a minimum length are eligible