Capacitors - long Q - full procedure, very grateful if someone will check

[pat666]

Homework Statement

Please see attachment.
I really need someone to check this properly please- I know this is meant to help people but checking this would be a huge help to me since its worth marks...(Ive done all the work)

THANKS A LOT to anyone who helps..

Homework Equations

The Attempt at a Solution

a) C=$$\epsilon$$₀*A/d
C=Q/V
0.1*10^-6=150*10^-6/V
V=1500V

C$$\propto$$ A
C$$\propto$$ 1/d

C=.1/4
C=.025uF

b)
q_max=C$$\epsilon$$
q_max=.01*10^-4C

q(t)=q_max*(1-e^-t/RC
7*10^-7/0.01*10^-4=1-e^{-(t/1*10^-3)}
t=1.2ms (just used solver for this)

c)
U=0.5CV^2
U=0.32J
$$\Sigma$$ C=6uF
U=0.5CV^2
0.32=6*10^-6V²
V=326.6V

 

[kuruman]

(a) OK
(b) The number is correct but you should learn to do it without solver. You may not have solver available if a question like this shows up on a test.
(c) Incorrect. What is the "difference" that you are asked to explain all about? Is potential energy stored in the capacitor(s) conserved or is there something else that is conserved?

 

[pat666]

thanks for the responce
b) i can solve this - just take ln both sides - but i don't have my calc with me so i used internet solver.
c) is the p.d i found correct? Yeah I just realized i forgot to do half the question. No potential energy can't be stored in the capacitor and that makes me think that my numerical answer is wrong to but I don't know how else to do it?
thanks again for responding.

 

[kuruman]

Try conserving charge instead of energy.

 

[pat666]

so qi=qf which I can see would be true as long as the system is closed which it is.
My problem is now that I can't find an appropriate formula anywhere in my textbook - online I found Q=CV which would give me
Qi=1.6*10^-3 C
therefore 1.6*10^-3= (2+4uf)V
so V=266.66V ??

for the energy can I use U=0.5CV^2 ?
and get
Ubefore = .5*4*10^-6*400^2 =0.32J ?
Uafter =.5*6*10^-6*266.66^2 =0.21333J

If that is correct (I doubt) than that proves that energy is not conserved, do you think that that is what they want explained?

 

[kuruman]

so qi=qf which I can see would be true as long as the system is closed which it is.
My problem is now that I can't find an appropriate formula anywhere in my textbook - online I found Q=CV which would give me

If your textbook does not have Q = CV (the definition of capacitance) you should toss it and get a better textbook.

Qi=1.6*10^-3 C
therefore 1.6*10^-3= (2+4uf)V
so V=266.66V ??

for the energy can I use U=0.5CV^2 ?
and get
Ubefore = .5*4*10^-6*400^2 =0.32J ?
Uafter =.5*6*10^-6*266.66^2 =0.21333J

I agree with these numbers.

If that is correct (I doubt) than that proves that energy is not conserved, do you think that that is what they want explained?

That's what they want you to explain, the apparent non-conservation of energy.

 

[pat666]

thankyou! just looking textbook does have it I am just an idiot -it was in the form C=Q/V (actually knew this of the top of my head just didn't realize its use.) so i flicked past it thinking it was useless to me... no harm
thanks

 

[pat666]

Hey kuruman, just one more question if you have time or see this.. with the explanation - I don't understand where the energy could have gone? How can energy not be conserved in a closed system?

 

[kuruman]

Hey kuruman, just one more question if you have time or see this.. with the explanation - I don't understand where the energy could have gone? How can energy not be conserved in a closed system?

Not all closed systems necessarily conserve mechanical energy which is what we are talking about here. Consider the totally inelastic collision where mass m₁ is moving with initial speed v₀ and collides with m₂ initially at rest. The two masses move together after the collision and their common speed is the speed of the center of mass, namely

$$V=\frac{m_1v_0}{m_1+m_2}$$

If you calculate the final energy of the "closed" system, you get (after a bit of algebra)

$$K_f=\frac{1}{2}m_1v^{2}_{0} \left( \frac{m_1}{m_1+m_2} \right)$$

This is clearly less than the initial energy

$$K_i=\frac{1}{2}m_1v^{2}_{0}$$

Where did the energy go? The stock answer is "Heat generated by friction."

This capacitor problem is exactly analogous to the totally inelastic collision. The symbols are different but the math is the same. Here is why

Momentum (p) is conserved becomes charge (q) is conserved.
Definition p = mv is replaced by definition q = CV, i.e. capacitance (C) is the analogue of mass (m) and voltage (V) is the analogue of velocity (v).
Kinetic energy=(1/2)mv² becomes Energy stored=(1/2)CV².

So the prediction is that the energy remaining in the system is the analogue of K_f, namely

$$E_f=\frac{1}{2}C_1V^{2}_{0} \left( \frac{C_1}{C_1+C_2} \right)$$

Put in the numbers and see what you get.

As for "where did the energy go?", complete the analogy. If heat production by friction accounts for loss of mechanical energy in a collision, what accounts for loss of mechanical energy in a circuit and how is it generated?

 

[pat666]

I know from empirical evidence that heat is a product of some reaction in most if not all circuits but i don't know why. building a TC in the holidays my caps where literally bursting from heat and although I had read a lot on the net about it happening to other people, I never understood where the heat was coming from? Having said that I guess that there is energy loss in other areas too? I can't imagine that a few resistors would produce much heat if they are rated for the use?

 

[kuruman]

http://en.wikipedia.org/wiki/Joule_heating