Charged Particle Oscillating in External Constant Electric Field


by scarletx09
Tags: differential eqn, electric field, osciallation
scarletx09
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#1
Feb4-12, 03:02 PM
P: 5
1. The problem statement, all variables and given/known data

A mass m that has net electric charge Q is oscillating along the x-direction on one end of a spring (whose other end is anchored) of relaxed length s0. Suppose that someone then swirches on an electric field E  that is uniform in space, constant in time, and which points in the +x direction. The entire system is then immersed in this electric field.

a. Set up the governing differential equation of motion for the mass in a coordinate system
with origin at the “anchored” end of the spring. Ignore gravity.

b. Without actually deriving it, what do you anticipate that the governing differential
equation in a “smart” coordinate system would be?

c. Go through steps analogous to those we went through in our text discussion to show that
the equivalent differential equation expressed in terms of the “right” variable is
independent ofE 

Hint: As part of this, you will need to find the equilibrium position with the field “on.”



Any idea of where to start would be greatly appreciated, thanks!
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tiny-tim
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#2
Feb4-12, 03:15 PM
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hi scarletx09!
Quote Quote by scarletx09 View Post
A mass m that has net electric charge Q is oscillating along the x-direction on one end of a spring (whose other end is anchored) of relaxed length s0. Suppose that someone then swirches on an electric field E  that is uniform in space, constant in time, and which points in the +x direction. The entire system is then immersed in this electric field.

a. Set up the governing differential equation of motion for the mass in a coordinate system with origin at the “anchored” end of the spring. Ignore gravity.
put mass times acceleration on the LHS, and all the forces on the RHS (as a function of x) …
what do you you get?
scarletx09
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#3
Feb4-12, 03:46 PM
P: 5
Using F=ma I got:

Net force = Force of the electric field + Force of spring
F = -QE - K(x-S0) = ma

(x-s0) = the displaced length from equilibrium

that resulted in:

a = (-QE -Kx + K(s0))/m

Does that seem correct? can i set up the equation of motion from this?

tiny-tim
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#4
Feb4-12, 03:53 PM
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Charged Particle Oscillating in External Constant Electric Field


Quote Quote by scarletx09 View Post
a = (-QE -Kx + K(s0))/m

Does that seem correct? can i set up the equation of motion from this?
(isn't it +QE ?)

yes, the question ask for a differential equation, so use d2x/dt2 instead of a

(for part b, you may need to use the "hint")
scarletx09
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#5
Feb4-12, 04:09 PM
P: 5
Thanks so much for your guidance with part a!

From my understanding of Part B:
the first equilibrium position was at s0 which led to that diff eq. and now its asking for a different (smater i guess?) diff. eq. based on a new equilibrium position with the electric field "on"

I am confused about how the equilibrium position would change with the electric field on. Didn't I take it into account in Part A?
tiny-tim
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#6
Feb4-12, 04:22 PM
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Quote Quote by scarletx09 View Post
From my understanding of Part B:
the first equilibrium position was at s0 which led to that diff eq. and now its asking for a different (smater i guess?) diff. eq. based on a new equilibrium position with the electric field "on"

I am confused about how the equilibrium position would change with the electric field on. Didn't I take it into account in Part A?
you need somehow to squeeze QE inside that bracket
Quote Quote by scarletx09 View Post
F = -QE - K(x-S0) = ma
scarletx09
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#7
Feb4-12, 04:48 PM
P: 5
ok. the equilibrium position would occur when Fspring = Fefield
--> -K(x-s0) = QE
(x-s0) = (QE)/-k
plugged that into my first diff. eq., the K cancelled and i got:

d^2t/dt^2 = (2QE)/m
tiny-tim
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Feb4-12, 04:55 PM
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Quote Quote by scarletx09 View Post
ok. the equilibrium position would occur when Fspring = Fefield
--> -K(x-s0) = QE
(x-s0) = (QE)/-k
yes that certainly works

but easier would be to rewrite F = -QE - K(x-S0)

as F = - K(x-S0 + QE/K)
plugged that into my first diff. eq., the K cancelled and i got:

d^2t/dt^2 = (2QE)/m
i'm confused


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