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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

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 ...


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 ...



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