AP Physics E&M MC: Electric Field, Induction, LC Circuit

[pietastesgood]

Homework Statement

Answer is B.

Answer is E.

Answer is E.

Homework Equations

E=F/q
E=V/d
Right hand rule for inductors

The Attempt at a Solution

Quite confused for these problems. For number 18, I'm quite baffled as to where the 0.04 meters even comes from. What I had attempted to do was find the potential at the point, 25 V, and divide it by the radius, or √(.15^2+.08^2), but that didn't even yield one of the answers. Then I realized that the potential was actually increasing as x increased, and that the object causing the electric field was likely off the graph. Then I just got really confused.

For number 33, I thought it was C. As the switch is flipped, there is a sudden increase in current in the top loop. This current flows out of the positive terminal, through the resistor, and back to the negative terminal. Since the current flows from left to right near the wire loop, the magnetic field generated should be into the page, right? Thus, to counteract this, a current that is counterclockwise will be generated in the loop. Unless I did the right hand rule wrong, that's what it should be. But the answer is E.

For number 35, I don't really understand what the question is asking. I've never learned about the frequency and oscillating of the current in a circuit, with just a LC circuit. Utterly confused here.

Would greatly appreciate any help! Thanks in advance!

 

[rude man]

Too many problems for one go!

For the ring problem: what is the direction of discharge current? So then what is the direction of the magnetic flux through the ring? Then either go with Lenz' law or Faraday's law to determine the direction of ring current. Lenz' is easier).

Is the discharge current increasing or decreasing with time?

 

[SammyS]

Homework Statement

[ IMG]http://imageshack.us/a/img405/2379/182008.jpg[/QUOTE]

Answer is B.

[ IMG]http://imageshack.us/a/img526/5937/332008.jpg
Answer is E.
[ IMG]http://imageshack.us/a/img203/9984/352008.jpg
Answer is E.

Homework Equations

E=F/q
E=V/d

The Attempt at a Solution

Quite confused for these problems. For number 18, I'm quite baffled as to where the 0.04 meters even comes from. What I had attempted to do was find the potential at the point, 25 V, and divide it by the radius, or √(.15^2+.08^2), but that didn't even yield one of the answers. Then I realized that the potential was actually increasing as x increased, and that the object causing the electric field was likely off the graph. Then I just got really confused.
...

Would greatly appreciate any help! Thanks in advance!

It's best not to put too many questions into one thread.

What radius are you referring to ?
How far are the 30V and 20V equi-potential lines apart in the vicinity of point P ?

 

[SammyS]

...

For number 35, I don't really understand what the question is asking. I've never learned about the frequency and oscillating of the current in a circuit, with just a LC circuit. Utterly confused here.

Would greatly appreciate any help! Thanks in advance!

What did you learn regarding LC circuits ?

 

[pietastesgood]

Oops, I'll make sure to spread the problems apart more next time. I understand number 18 now. E=-dV/dr, so the strength of the electric field is the change in voltage/change in distance. So since the equipotential lines for 20 and 30 V are .04 m apart, it would be a change of 10 V/a change of .04 m, so B.

For the Lenz' law problem, the direction of discharge out the positive terminal and along the right on the bottom of the circuit. Direction of magnetic flux is downward through the ring. Ring will attempt to counteract that, so it creates magnetic flux upward. According the to RHR, that makes the current counterclockwise, which makes me confused since the answer is clockwise. Did I have a logical error somewhere? Discharge current decreases over time.

Regarding LC circuits, I only learned that at t=0, an inductor in a circuit acts like a broken wire, while at t=∞ it acts like a wire.

 

[rude man]

For the Lenz' law problem, the direction of discharge out the positive terminal and along the right on the bottom of the circuit. Direction of magnetic flux is downward through the ring.

Direction of flux is INTO the ring (into the page).

Ring will attempt to counteract that, so it creates magnetic flux upward. According the to RHR, that makes the current counterclockwise, which makes me confused since the answer is clockwise. Did I have a logical error somewhere? Discharge current decreases over time.

EDIT: Nope, I'm wrong. Stay tuned for an explanation.

 

[rude man]

Assume for the moment the B field due to the wire is increasing. That would generate a B field in the ring to oppose the buildup of total B in the ring, so that current would be ccw. But the wire current is decreasing, so the ring current will be cw to try to maintain the total ring B field unchanged. In other words, the ring current opposes the direction of change of wire B, which is negative, not B itself. So the answer is indeed E.

 

[pietastesgood]

Aha! That makes sense. Thank you! Now, just the oscillator problem.

 

[rude man]

Aha! That makes sense. Thank you! Now, just the oscillator problem.

OK, if you don't know the formula for the oscillating frequency of an L-C oscillator you'll have to derive it!

Connect a C and an L together, put an initial charge on C and solve the ODE for V(t)!

 

[pietastesgood]

I got some of it after looking up a video on Youtube.

ε=Q/C for a capacitor

Q/C - L(dI/dt) = 0
Q/C - L(-d^{2}q/dt^{2}) = 0
d^{2}q/dt^{2} = -Q/LC

However, I don't quite follow how the angular frequency ω=1/\sqrt{LC} from that second ODE.

 

[rude man]

I got some of it after looking up a video on Youtube.

ε=Q/C for a capacitor

Q/C - L(dI/dt) = 0
Q/C - L(-d^{2}q/dt^{2}) = 0
d^{2}q/dt^{2} = -Q/LC

However, I don't quite follow how the angular frequency ω=1/\sqrt{LC} from that second ODE.

Stick q = q0*cos[t/√(LC)] into your last equation, what do you get?
(q0 is the initial charge on C).

 

[pietastesgood]

d^2q/dt^2 = -q0*cos[t/√(LC)]/LC

I'm only in Calculus II, so I don't really follow how to solve the second order differential equation.

 

[rude man]

d^2q/dt^2 = -q0*cos[t/√(LC)]/LC

I'm only in Calculus II, so I don't really follow how to solve the second order differential equation.

Pleae look at post #11 again. You don't have to solve it. I made the assumption that you didn't know how.

 

[pietastesgood]

Sorry, but I'm not sure what else I can do with plugging in q = q0*cos[t/√(LC)] into the second order differential equation, other than replacing -Q/LC with -q0*cos[t/√(LC)]/LC.

 

[rude man]

Sorry, but I'm not sure what else I can do with plugging in q = q0*cos[t/√(LC)] into the second order differential equation, other than replacing -Q/LC with -q0*cos[t/√(LC)]/LC.

First of all, there is only one q. Don't call it q one time and Q another.

If q = q0*cos[t/√(LC)] what is d²q/dt²?

Now, put that into your equation. What do you get? An identity, perhaps?

 

[pietastesgood]

Well, taking the second derivative of q0*cos[t/√(LC)] yields -(q0/(LC))*cos(t/√(LC))

Setting that equal to -q/LC, simplifying, I get q0cos(t/√(LC))=q again. Pretty sure I didn't reach the right conclusion there.

 

[rude man]

Well, taking the second derivative of q0*cos[t/√(LC)] yields -(q0/(LC))*cos(t/√(LC))

Setting that equal to -q/LC, simplifying, I get q0cos(t/√(LC))=q again. Pretty sure I didn't reach the right conclusion there.

Yes. You got an identity which means the solution for q that I gave you is a correct solution.

Now, if someone gives you a signal cos(t/√(LC) do you think you can figure out what the frequency of that signal is?

 

[pietastesgood]

Oh, right. cos(wt)=cos(t/√(LC)), so w=1/√(LC)
There we go!