Calculating Charge on Capacitor with Decreasing Magnetic Field & Coil

[John123]

Homework Statement

A vertical magnetic field decreases at a rate of 94.0 mT/s. A flat circular coil of 810 turns with a 41.0 cm diameter has a 153 uF capacitor across its terminals. If the coil is placed perpendicular to the field.

What is the charge on the capacitor?

Relevant Equations

q = CV

This is where I'm stuck, I have a few equations, but none of them use the right variables of:
R = 20.5cm
C = 153uF
e(the one per time) = 94000T/s
N = 810

Where should I start?

 

[kuruman]

You should start by understanding the physical process that takes place. Why should there be any charge on the capacitor? Hint: Q = CV says that if there is no V, there is no Q. So where does the V come from?

 

[rude man]

You should start by understanding the physical process that takes place. Why should there be any charge on the capacitor? Hint: Q = CV says that if there is no V, there is no Q. So where does the V come from?

I think I disagree that there is no $V_c$.
Reasoning:
1. There is an induced field $E_m$ around the coil such that $2πaNE_m$ = emf where a = coil radius, N = no. of turns
and emf = −Ndϕ/dt.
2. Current i must be zero since we can't have $d(V_c)/dt$ buildup over time.
3. But the net E field in the wire = 0
4. So a static field $E_s$ must exist in the coil wire = $−E_m$.
5. but the circulation of $E_s = 0$.
6. So an equal & opposite $E_s$ must exist across the capacitor.
7. Therefore the V across C = $V_c = 2\pi aE_s$ .

 

[kuruman]

I think I disagree that there is no $V_c$.
Reasoning:
1. There is an induced field $E_m$ around the coil such that $2πaNE_m$ = emf where a = coil radius, N = no. of turns
and emf = −Ndϕ/dt.
2. Current i must be zero since we can't have $d(V_c)/dt$ buildup over time.
3. But the net E field in the wire = 0
4. So a static field $E_s$ must exist in the coil wire = $−E_m$.
5. but the circulation of $E_s = 0$.
6. So an equal & opposite $E_s$ must exist across the capacitor.
7. Therefore the V across C = $V_c = dE_s$ where d is the capacitor effective gap.

All that is good, but your detailed description gave away what I was attempting OP to see on his own. I never said there is no voltage across the capacitor.

You should start by understanding the physical process that takes place. Why should there be any charge on the capacitor? Hint: Q = CV says that if there is no V, there is no Q. So where does the V come from?

The hint clearly states "... if there is no V there is no Q ..." The hypothetical statement is undoubtedly true and does not assert that there is no voltage; it just describes one (trivial) aspect of the equation. If OP were to couple this trivial statement with the task to find Q, then OP might understand the necessity to seek the source of V, a direction I pointed at by asking, "Why should there be any voltage across the capacitor?" Perhaps I was too subtle ...

 

[rude man]

All that is good, but your detailed description gave away what I was attempting OP to see on his own. I never said there is no voltage across the capacitor.

The hint clearly states "... if there is no V there is no Q ..." The hypothetical statement is undoubtedly true and does not assert that there is no voltage; it just describes one (trivial) aspect of the equation. If OP were to couple this trivial statement with the task to find Q, then OP might understand the necessity to seek the source of V, a direction I pointed at by asking, "Why should there be any voltage across the capacitor?" Perhaps I was too subtle ...

I guess I was too hasty but I had the definite impression that your hint was for 'zero voltage'. You did bold out any in that sentence. OK. Mea culpa.
PS it's only very recently that the concept of duality of E fields was even introduced anywhere in PF as far as I know so I have my doubts that the OP would have hit on it. I think I left a good bit to him/her to contemplate even before attempting the numerical solution.