The end point is at $R/2$, and for the potential we need only consider the charge inside the sphere of radius $R/2$. The charge inside this distance is $Q/8$ so the final potential is $V_2 = kqQ/8 (R/2) = kqQ/(4R)$.
@JohnRennie why is the rms voltage from the emf source not equal to the sum of the rms voltage across each of the circuit elements in a series RLC circuit?
ah, but is it true then that the emf source as a function of time is equal to the sum of the voltages across each of the circuit elements as a function of time?
@Aladdin the equilibrium position is where the net force on the particle is zero. Since the force is $dV/dx$ this corresponds to a minimum in the (total) potential energy.
@Aladdin You're assuming that the maximum in the potential energy is at the origin, and this may not be the case.
In fact, since the question asks you for the KE at the origin the maximum is almost certainly not at the origin otherwise the answer would simply be zero.
@pi-π it's a slightly odd question. I think you're supposed to imagine the electron is moving in a ring and then calculate the torque on the ring.
If the magnetic field is normal to the plane of the ring then the torque on the ring is zero because the Lorentz force always acts towards the centre of the ring. However in this case the ring has been tilted at an angle to the field so there will be a torque.
Shall I draw a diagram to illustrate what I think the question is asking?
OK, I think this is the situation the question is describing:
The ring is now tilted by 30°
And there will now be a torque on the ring.
The way I'd probably do this is that the electron moving in a ring creates a dipole $\mu$ and the torque on a dipole $\mu$ in a field $B$ is $\tau = \mu \times B$.
$\mu \times B = \mu B \sin30 = \tfrac{1}{2}\mu B$
And the dipole $\mu$ will be given by the usual expression for a current loop.
I'd probably start by shifting the coordinates to put the dipole at the origin. If you add the displacement (-2, 3, 1) to both positions this puts the dipole at (0, 0, 0) and the point B at (2, 2, -1). OK so far?
The diagram on the left shows the situation described by the question, but I would redraw it to put the dipole at the origin as I've done on the right.
A particle of mass m, charge -Q is constrained to move along the axis of ring of radius a. The ring carries a uniform charge density +λ along its circumference. Initially the particle lies in the plane of the ring where no force acts on it. The period of oscillation if it's displaced slightly from its equilibrium position is
In this question, initially the particle is at center then displaced right?
The potential is quite a complicated function of the distance along the axis $x$, but very near to the centre of the ring we can approximate it as $V(x) \approx \tfrac{1}{2} k x^2$ for some constant $k$ that we have to work out.
We'll get this constant by using a binomial expansion and keeping only the first term.
It doesn't matter whether we use potential or force because either way we want the value of the constant. I used the potential because I find it easy to remember the equation for the potential of a charged ring. You can work with the force if you prefer.
Because you're looking from above you can't see the vertical component of the normal force. You see only the horizontal component that I've marked in red.
When the rails are parallel the two horizontal components are equal and opposite so the cancel out and the cones don't move.
But with the rails angled the horizontal components of the normal force no longer cancel. There is a component upwards on the diagram so the cone will move up.
@JohnRennie I was directed to this site by someone. My question is this. I welded aluminum to steel rebar. The steel has carbon and iron in it. I took my multimeter and connected 1 prong to the steel and another on the aluminum. I got 0.6 volts from it. How does this work?
Problem Solving Strategies
Problem Solving Strategies