Recent content by Jasso

Taken together, the two different paths between A and B form a closed loop. To see this, go along P1 from A to B, then go along P2 from B to A. If the integral along this loop is zero, then that means that the integral of P1 from A to B is equal and opposite to the integral of P2 from B to A...

What would the field look like around an infinitely short current segment dl? Would it still form concentric rings, or would it be different?

The electric field lines are a representation of the direction of force that a test charged particle would experience. In your example, the 2nd positive charge would still be repelled from the 1st positive charge and attracted to the negative charge.

One of the biggest things that gets me is the concept of relativistic mass, i.e. that mass increases at relativistic speeds. It's led to more than a few misunderstandings on these forums, mostly from thinking that a fast enough particle will turn into a black hole because of it.

You are making the assumption that you are grounded. Pretty much the only way you can have the charge flow through your body to the Earth is if your body was in contact with a grounded conductor.

They are two different properties that have the same effect. Your first source, which discusses the covalent bond, is generally talking about molecules, while the other source is talking about individual atoms. If a molecule has a strong covalent bond, it generally makes a good insulator because...

What is the magnitude of the force from S1 to S2, and what is the magnitude of the force from S2 to S1?

I believe the beginning they are talking about a chemical battery; it talks about moving the charge from a low energy state to a high energy state. In a battery, a chemical reaction supplies the energy to separate the charges between the cathode and anode, which causes an electrical potential...

You have already been given answers by the others in this thread. They have told you that moving neutral matter does not generate magnetic fields, and gave you a reference to where you can learn more about magnetism. You then go on to argue with the people you are asking help from. That is not...

It's not that you can't use the divergence theorem, as it holds no matter the case. You cannot use cylindrical symmetry to simplify the system because the E field in the system will not be cylindrically symmetric. To prove this to yourself, take the shape of the field when z>>L. Even as a...

There would still be an induced momentum. As the magnetic field comes through, it would be a time varying field at the surface. Since -dB/dt = curl(E), there would be an equivalent electric field at the surface pushing the charged shell the other way.

This won't change the fact that the momentum of the sphere changes; it would only mean that the momentum of the sphere is indeterminate. But the magnetic field generated by the moving charge inside is not spherically symmetric, it curls around the axis of motion of the moving charge. The...

You aren't measuring the velocity, you are measuring the radius of curvature of the beam. When you find the curvature, you plug that in, along with other values into an equation to find the e/m. For a given velocity and magnetic field strength, the e/m is given by e/m = v / (r B). So as the...

Let me elaborate a little on what jtbell said. As the particle passes through the capacitor (let's assume that it enters at the exact midpoint between the plates), it will be deflected to the side. As it moves closer to one plate, the electric potential will drop because it will be farther from...

I think you may be over-thinking it. If df(t)/dt = C, then f(t) can be written as f(t) = f(0) + Ct, i.e. a first order Taylor expansion. That means that f(t) = <A>t can also be written as <A>t = <A>0 + d/dt(<A>t) * t.