Charge of Capacitor in LC Circuit

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Homework Statement



An LC circuit constructed of a 1 microF capacitor and a 1 microH inductor
is set in oscillation so that the charge on the capacitor is 1.0 microC at t = 0
and 2.0 microC at t = 1.57 x 10^-6 s. What is the charge on the capacitor at
t = 1.0 s?

Homework Equations



Q(t) = Q0cos(wt), w = angular frequency = 1 / sqrt(LC), L = inductance

The Attempt at a Solution



At first glance, it seemed like a simple plugin problem using t = 1.0 s and Q0 = 1.0 microC in the equation..

but i'm not sure why the problem tells me the charge at t = 1.57 x 10^-6 s

is this just extra unrelated information? or is my method incorrect?
 

Answers and Replies

  • #2
G01
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The problem is that Q=1microC is not the amplitude of the charge oscillation.

You are given the charge at two different times. You need to use this to find [itex]Q_o[/itex]
 
  • #3
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ok, but i am not sure how to do this..

i thought Q0 was the initial condition charge, and it says at t = 0, the charge is 1.0 microC.. Isn't this the initial condition?
I have no idea what to do with the other charge at the other time..
 
  • #4
collinsmark
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ok, but i am not sure how to do this..

i thought Q0 was the initial condition charge, and it says at t = 0, the charge is 1.0 microC.. Isn't this the initial condition?
It is an initial condition, technically yes. But as G01 points out, it doesn't necessarily mean that Q = Q0 at time t = 0.

In general (not necessarily specific to this problem), Q can be anywhere between +Q0 and -Q0 at time t = 0, depending on how t0 is defined. Q doesn't necessary have to start at maximum amplitude.
I have no idea what to do with the other charge at the other time..

Q = Q0cos(ωt + θ)​
You've already expressed the formula for ω. Use your two initial conditions to solve for Q0 and θ. :wink:

[Edit: In case you were wondering where the t0 fits in, the above approach is equivalent to saying
Q = Q0cos(ω[t - t0]),
and then use your initial conditions to solve for Q0 and t0. You'll get the same final answer either way. :smile:]
 
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  • #5
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ok, I am still a little confused, since in the lecture, I did not learn these two equations:

Q = Q0cos(ωt + θ)
Q = Q0cos(ω[t - t0]),

I was told that Q(t) = Acos(wt) + Bsin(wt), and that A and B depends on initial conditions

Then, for initial conditions at t = 0, A = Q0, and B = 0, giving me
Q(t) = Q0cos(wt)

But I do not know how to use the two times and two charges..there are no examples like this in the textbook or in the lectures..
I think i see what you are saying..I am supposed to write two equations from the two initial conditions, and solve a system of equations... but could you express this the way i learned it with the cos(wt) and sin(wt) ?
 
  • #6
collinsmark
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ok, I am still a little confused, since in the lecture, I did not learn these two equations:

Q = Q0cos(ωt + θ)
Q = Q0cos(ω[t - t0]),

I was told that Q(t) = Acos(wt) + Bsin(wt), and that A and B depends on initial conditions

Then, for initial conditions at t = 0, A = Q0, and B = 0, giving me
Q(t) = Q0cos(wt)

But I do not know how to use the two times and two charges..there are no examples like this in the textbook or in the lectures..
I think i see what you are saying..I am supposed to write two equations from the two initial conditions, and solve a system of equations... but could you express this the way i learned it with the cos(wt) and sin(wt) ?
Actually, that works just as well too. As it works out,

Q = Q0cos(ω[t - t0]) = Q0cos(ωt + θ) = Acos(ωt) + Bsin(ωt).​

So if you'd rather, use

Q = Acos(ωt) + Bsin(ωt)​

if you want. That's completely fine. :approve: No matter which of the equations you pick, it all still works. They're all just different ways of expressing the same thing.

Plug in (t = 0, Q = 1.0 μC), and solve for A. Once you know A, plug in (t = 1.57 x 10-6 sec, Q = 2.0 μC) and solve for B.
 
  • #7
I apologize if this isn't relevant to the question, but I'm working on a similar situation where I get the Qocos(wt) term from the particular solution of my differential equation, but why is the complementary solution not included in the entire expression for Q(t)? Specifically, I have the term Qocos((LC)-1/2t) which has the opposite sign of the particular solution? Any insight as to what to do with the complementary solution?

Actually, that works just as well too. As it works out,

Q = Q0cos(ω[t - t0]) = Q0cos(ωt + θ) = Acos(ωt) + Bsin(ωt).​

So if you'd rather, use

Q = Acos(ωt) + Bsin(ωt)​

if you want. That's completely fine. :approve: No matter which of the equations you pick, it all still works. They're all just different ways of expressing the same thing.

Plug in (t = 0, Q = 1.0 μC), and solve for A. Once you know A, plug in (t = 1.57 x 10-6 sec, Q = 2.0 μC) and solve for B.
 
  • #8
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i'm not sure what you mean by complementary solution...

but for my problem, i solved for A = 1 x 10-6 C, and B = -6.37 x 1011

and then I did

Q(t) = 1 x 10-6 cos(wt) + (-6.37 x 1011) sin(wt)
= 2.23 x 1011 C

is this answer correct? am i supposed to get an enormous charge at 1 second?
 
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  • #9
i'm not sure what you mean by complementary solution...

but for my problem, i solved for A = 1 x 10-6 C, and B = -6.66 x 1011

and then I did

Q(t) = 1 x 10-6 cos(wt) + (-6.66 x 1011) sin(wt)
= 2.33 x 1011 C

is this answer correct? how come i get an enormous charge at 1 second?

I have a cosinusoidal voltage source with a series LC circuit which results in an inhomogeneous differential equation that I solved via the method of undetermined coefficients.
 
  • #10
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sorry, I am only in a general physics with Calculus II class...so I am not sure how to help..

can someone confirm if my answer is correct?
 
  • #11
You may want to check again for B. I get B = 2*10-6, approximately. Since the magnitude of sin(wt) is at most 1, does a value for B larger than Q make sense?

sorry, I am only in a general physics with Calculus II class...so I am not sure how to help..

can someone confirm if my answer is correct?
 
  • #12
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oops, i calculated wt and A wrong...
now i get B = 1.99x10-6 and
Q(t) = (1 x 10-6) cos(wt) + (1.99 x 10-6) sin(wt)

Q(1) = 2.37 x 10-7 C

Is this the correct answer for the problem?
 
  • #13
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I used both methods and got Q(1) = 0.536 C.
For Schecter5150, what kind of differential equation are you working with?
 
  • #14
I used both methods and got Q(1) = 0.536 C.
For Schecter5150, what kind of differential equation are you working with?

I have the equation:

L*d2Q/dt2 + Q/C = Eocos(wt)

And I just realized that the root of my auxiliary equation is r = j/sqrt(LC), which means my complementary solution would be of the form:

Qc(t) = c1*cos(rt) + c2*sin(rt), which is entirely imaginary, correct? In that case, could I neglect the complementary solution and only consider the particular, which I have computed to be:

Qocos(wt)
 
  • #15
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I have the equation:

L*d2Q/dt2 + Q/C = Eocos(wt)

And I just realized that the root of my auxiliary equation is r = j/sqrt(LC), which means my complementary solution would be of the form:

Qc(t) = c1*cos(rt) + c2*sin(rt), which is entirely imaginary, correct? In that case, could I neglect the complementary solution and only consider the particular, which I have computed to be:

Qocos(wt)

By "root of my auxiliary equation", do you mean:
[tex]Q_{c}=c_{1}e^{\frac{j}{\sqrt{RC}}}+c_{2}e^{\frac{-j}{\sqrt{RC}}}[/tex] ?
Because if that's the case, make sure you apply Euler's formula correctly.
 

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