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GR9677 #4
Problem
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Electromagnetism$\Rightarrow$}Small Oscillations

The potential is determined in the previous problem to be $V=\frac{Q}{4\pi \epsilon_0 \sqrt{R^2+x^2}}$. The field is given by $\vec{E}=-\nabla V$. Before taking derivatives, one can simplify the potential since it is given that $R >> x$.

Binomial expand it ($(1+y)^n\approx 1+ny$, for $y$ small) to get

$V\approx \frac{Q}{4\pi \epsilon_0 R}(1+\frac{x^2}{2R^2})
$

Taking the derivative, using the equations $\vec{E}=-\nabla V$, and $\vec{F}=q\vec{E}$, one gets,

$-q\frac{Q}{4\pi \epsilon_0 R}\frac{x}{R^2}=F=m\ddot{x}.
$

Small oscillations have the same form as simple harmonic oscillations, i.e., $\ddot{x} = -\omega^2 x$. The angular frequency is $\omega=\sqrt{q\frac{Q}{4\pi \epsilon_0 R^3 m}}$, as in choice (A).

Alternate Solutions
 djh1012014-09-19 21:38:40 To expand a little bit on the quick and dirty method (point 2, specifically): 1. Small oscillations aren't going to depend on displacement. Eliminate BDE. 2. Experience with small oscillations should tell you that from the potential you are going to get a Hooke's-Law-esgue force, which will give you k. To get the frequency, take the square root of k and divide by the square root of m. Only A has root m in the denominator.Reply to this comment archard2010-05-14 10:06:55 The way I did it is to remember from freshman physics that the period of an oscillation (and thus the frequency) is independent of displacement. That eliminates B, D and E. A and C are the same save for a square root, so one of them has incorrect dimensions. Choose the one with dimensions of frequency, A.Reply to this comment chri5tina2006-11-27 05:11:24 It is possible to do this problem w/o the binomial expansion. Take F = -k*x, plug in -dV/dx for F and solve for k. Plug k into w = (k/m)^1/2 and you'll have an expression with (R^2 + x^2)^3/2 in the denominator. At this point you can remember that R >> x and simply cancel out x as negligible, which leaves R^3 in the denominator and the correct solution.Reply to this comment
djh101
2014-09-19 21:38:40
To expand a little bit on the quick and dirty method (point 2, specifically):
1. Small oscillations aren't going to depend on displacement. Eliminate BDE.
2. Experience with small oscillations should tell you that from the potential you are going to get a Hooke's-Law-esgue force, which will give you k. To get the frequency, take the square root of k and divide by the square root of m. Only A has root m in the denominator.
maxdp
2013-09-24 13:55:12
It seems stiner905's comment hasn't been noted since s/he didn't mark it correctly, so I'll repost it.

"After the binomial expansion, the potential should have a "-" instead of a "+" in the expansion term. However, a minus exists in the force equation because the test charge has charge -q."

In other words, there are two sign errors in the current solution that cancel out.
shak
2010-08-13 15:02:04
The best and easiest way to solve this problem is using Lagrangian equation of motion.. it is one dimensional system.. so small charge -q is moving only along the x-axis.
Lagrangian is
L=E- V (1)
and
E=m*x(dot)^2/2 (2)
V= same potential in problem 3.
archard
2010-05-14 10:06:55
The way I did it is to remember from freshman physics that the period of an oscillation (and thus the frequency) is independent of displacement. That eliminates B, D and E. A and C are the same save for a square root, so one of them has incorrect dimensions. Choose the one with dimensions of frequency, A.
 nkqed2011-10-24 17:01:10 I'm sorry but this is the wrong way to think about it. x is not the displacement. We are displacing the particle perpendicular to x. And if you consider the situation where R is not >> x then we will have some dependence on x.
 ian2012-10-05 09:56:53 nkqed: x is the displacement - look at the diagram. Also, the problem only asks us to consider the case where R >> x. I don't see anything wrong with archard's approach.
 Prufrock2013-09-16 14:44:53 This is a good way to approach it. Barring possible pathological cases, the frequency in a SHO "obviously" doesn't depend on the displacement.
Muphrid
2009-10-08 14:06:53
It may be faster to keep working with electrical potential to find the spring constant. Simple harmonic oscillators satisfy $F(x) = -kx$, as has been said, but they also satisfy

$U(x) \approx U_0 + \frac{1}{2} k x^2$

and have angular frequency

$\omega = \sqrt{\frac{k}{m}}$

It's important to remember that what you know from problem (3) is $V(x)$, potential energy per unit charge, not $U(x)$, which is real potential energy. This is what alleviates the sign confusion later on.

As in the base solution, you can binomial expand $V(x)$ and approximate it as

$V(x) = \frac{Q}{4\pi \epsilon_0} \frac{1}{R} \left( 1 - \frac{1}{2} \frac{x^2}{R^2} \right)$

Note that the 1 can be dropped; this just gives a constant overall potential that we're not interested in. This leaves us with

$V(x) = V_0 - \frac{1}{2} \frac{Q x^2}{4 \pi \epsilon_0 R^3}$

And now multiply through by $-q$, the charge of the test charge, to get real potential energy:

$U(x) = -q V_0 + \frac{1}{2} \frac{qQ x^2}{4 \pi \epsilon_0 R^3}$

Recognize this as a simple harmonic potential, where

$\frac{1}{2} k x^2 = \frac{1}{2} \frac{qQ x^2}{4 \pi \epsilon_0 R^3} \rightarrow k = \frac{qQ}{4 \pi \epsilon_0 R^3}$

And thus

$\omega = \sqrt{\frac{k}{m}} = \sqrt{\frac{qQ}{4 \pi \epsilon_0 m R^3}}$
tinytoon
2008-11-07 14:48:10
Also, you could derive the electric field from first principles to get the answer (although very inefficient). This is, nonetheless, a valid alternative:

$E = \int \frac{dq}{4\pi\epsilon_0 r^2}$

We only care about the horizontal component because the force in the vertical component is zero:

$E = \int \frac{dqcos\theta}{4\pi\epsilon_0 r^2}$.

Using $cos\theta = \frac{x}{r} = \frac{x}{\sqrt{x^2+R^2}}$ and $r^2 = x^2+R^2$, we get:

$E = \frac{qx}{4\pi\epsilon_0 (x^2+R^2)^{3/2}}$

and

$F = \frac{qQx}{4\pi\epsilon_0 (x^2+R^2)^{3/2}} =kx$

We can clearly see now that:

$k = \frac{qQ}{4\pi\epsilon_0 (x^2+R^2)^{3/2}}$.

We know that:

$\omega = \sqrt{k/m} = \sqrt{\frac{qQ}{4\pi\epsilon_0 (x^2+R^2)^{3/2}m}}$

In the limit that $R$>>$x$ , this reduces to:

$\omega = \sqrt{k/m} = \sqrt{\frac{qQ}{4\pi\epsilon_0 R^3m}}$.

gn0m0n
2008-10-19 17:49:35
What are we expanding, exactly?
dcan
2008-04-09 16:32:33
Can someone tell me why it doesn't work to use conservation of energy? Supposedly the potential energy is qV, but when I solve for $\omega$ I'm off by a factor of 1/$\sqrt{2}$.
Gaffer
2007-10-26 14:12:55
This problem has some particularly generous writers.

For small oscillations, I believe it is safe to say $\omega$ must be a constant, that is not dependent upon x. This knocks off BDE. Then all you need to remember is that $\omega$ will most likely involve a square root due to the diffEQ yosun mentioned and you have choice A, no dimensional analysis or calculation necessary.

Though of course it is always better to know why a particular formula is correct, sometimes quick and dirty is the way to go.

Oh that they were all this easy!
 jmason862009-09-29 16:40:35 This is exactly how I did the problem too. Definitely the right strategy under time pressure.
rmyers
2006-11-29 16:16:57
This can also be done just by dimensional analysis. Qq/(4pi(e0)R^2) has units of force = kg * m/s^2 and angular frequency is just 1/s. Only A & B are possible answers after this. From there I suppose you can just use the idea that the angular frequency usually doesn't depend on initial displacement in cases like these.
 eliasds2008-08-18 00:58:43 I believe that choice E has the same dimensions as A&B.
 eliasds2008-08-18 01:43:44 I believe that choice E has the same dimensions as A&B.
 wangjj01202008-10-11 01:12:02 rmyers: your comment "you can just use the idea that the angular frequency usually doesn't depend on initial displacement" is not always true. You can find the x dependence from the exact solution.
chri5tina
2006-11-27 05:11:24
It is possible to do this problem w/o the binomial expansion.

Take F = -k*x, plug in -dV/dx for F and solve for k.

Plug k into w = (k/m)^1/2 and you'll have an expression with (R^2 + x^2)^3/2 in the denominator.

At this point you can remember that R >> x and simply cancel out x as negligible, which leaves R^3 in the denominator and the correct solution.
stiner905
2006-10-29 12:26:06
After the binomial expansion, the potential should have a "-" instead of a "+" in the expansion term. However, a minus exists in the force equation because the test charge has charge -q.

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