Electric Field/Oscillation Problem

1. Jan 23, 2005

barnsworth

A small bead of mass m, carrying a charge q, is constrained to slide along a thin rod of length L. Charges Q are fixed at each end of the rod.

a) Obtain an expression for the electric field due to the charges Q as a function of x, where x is the distance from the midpoint of the rod. (Answer is kQ[(L/2+x)^-2 - (L/2-x)^-2]^I)

b) Show that for x << L, the magnitude of the field is proportional to x.

c) Show that if q is of the same sign as Q as, the force that acts on the object mass m is always directed toward the center of the rod and is proportional to x.

d) Find the period of the oscillation of the mass m if it is displaced by a small distance from the center of the rod and then released. (Answer is 2 * pi * sqrt(mL^3 /32kqQ) ).
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So part (a) was pretty easy. I just did E = kQ/r^2 – kQ/r^2 and got the book answer.

The next three parts i don't know what to do. For part (d) i know that T = 2 pi * sqrt(m / k).
Then i did kx = qE to solve for "k" of spring, plugged in E, but it left me with some overly complicated equation.

Anyways, thanks for any help in advance.

2. Jan 23, 2005

futb0l

For b), you got the equation so it will be pretty simple to explain that the magnitude is proportional of the E-field to x. For example, you can try and make x, 2 times larger, and see what happens. And remember that x << L.

c) Use coulomb's law and get the force.

d) Put all the forces together in one equation and solve for the period. Just like a normal Simple Harmonic Oscillation, I guess.

3. Jan 23, 2005

barnsworth

Thanks for the response, f -- but i'm still not quite sure not to do on (d).

I did:

Coulomb's law:
kQq/(1/2L+x)^2 - kQq/(1/2L-x)^2 = kx = ma.
a = kQq/m(1/2L+x)^2 - kQq/m(1/2L-x)^2.
2pi*f = sqrt(kQq/m(1/2L+x)^3 - kQq/m(1/2L-x)^3)

Which is definately wrong... I guess it would help if i remembered exactly how to do SHM.