Electric Field/Oscillation Problem

[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.

 

[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.

 

[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 definitely wrong... I guess it would help if i remembered exactly how to do SHM.