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Capacitors with dielectric oil

  • Thread starter scholio
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Homework Statement



a 5 microfarad capactior and a 10 microfarad capacitor are connected in parallel across a 50 volt battery as shown. the battery is disconnected and the two capacitors are separated from one another without altering the charge on either one. oil with a dielectric constant of 4 is inserted into the gap of the 5 microfarad capacitor, then the capacitors are reconnected.

what is the final potential across the parallel combination assuming:
_________ ____________
_|_ _|_ _|_ 50volts

_a _ _b_ _
|________|____________|

a = 5micro farads
b = 10 microfarads

a) the positive side of the each capacitor are connected together?
b) one capacitor was flipped so that the positive side of one was connected to the negative side of the other?

Homework Equations



parallel C = C_1 + C_2 = Q/V + Q/V

Q = CV where C is capacitance, V is electric potential, Q is charge

parallel plate capacitor with dielectric
C = kappa[(epsilon_0*A)/d] where kappa is dielectric constant, in this case = 4, epsilon_0 is constant = 8.85*10^-12, A is area of plate, d is distance between plates


The Attempt at a Solution



since size is not specified, it is assumed area A is constant
i initially thought i would sum the two capacitor in parallel to get 15 microfarads and then sub that in for C in the last eq, along with 4 for kappa, but then came across distance d which is not given. also that equation does not involve the voltage from the battery.

then i tried using the second equation for charge Q where i summed the Q's for each capacitor getting 2.5*10^-4 + 5*10^-4 = 6.5*10^-4, after getting Q net value i was unsure about how to include kappa = 4

this is where i am at, not sure about the equations nor whether my attempts were along the right track.
 
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Answers and Replies

  • #2
Doc Al
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Try to answer these questions:

(1) What charge is on each capacitor when they are first separated?

(2) What stays the same when the dielectric is inserted? What changes?

(3) When the two capacitors are connected, what can you say about the voltage across each? What happens to the charge on each?
 
  • #3
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1) charge on each capacitor using energy U = 1/2CV^2 where C is capacitance, V is electric potential

for the 5 microfarad capacitor --> U = 1/2 (5*10^-6)(50)^2 = 6.25*10^-3 joules
for the 10 microfarad capacitor --> U = 1/2(10*10^-6)(50)^2 = 1.25*10^-2 joules

2) adding the dielectric oil reduces (changes) the electric field and the electric potential difference in the 5 microfarad capacitor.
i am not sure about what remains the same, is it the capacitance of the 5 microfarad capacitor that remains the same?

3) i'm not sure about this, but when two capacitors are connected the voltage decreases from one and increase on the other capacitor, do they equalize? as for the charge, the charge decreases on one and increases the same amount on the other capacitor.

thanks
 
  • #4
Doc Al
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1) charge on each capacitor using energy U = 1/2CV^2 where C is capacitance, V is electric potential

for the 5 microfarad capacitor --> U = 1/2 (5*10^-6)(50)^2 = 6.25*10^-3 joules
for the 10 microfarad capacitor --> U = 1/2(10*10^-6)(50)^2 = 1.25*10^-2 joules
To find the charge, use the definition of capacitance: C = Q/V.

2) adding the dielectric oil reduces (changes) the electric field and the electric potential difference in the 5 microfarad capacitor.
Good.
i am not sure about what remains the same, is it the capacitance of the 5 microfarad capacitor that remains the same?
No. Introducing the dielectric increases the capacitance. What happens to the charge?

3) i'm not sure about this, but when two capacitors are connected the voltage decreases from one and increase on the other capacitor, do they equalize?
Absolutely. After the charge rearranges, the voltages must be equal.
as for the charge, the charge decreases on one and increases the same amount on the other capacitor.
Good. The total charge doesn't change.

Make use of this to figure out the final voltage in each case.
 
  • #5
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so the dielectric oil increases the capacitance and the charge increases too using C = Q/V and solving for Q

the charge for the 5 microfarad capacitor is Q = C_1V = 5*10^-6(50) = 2.5*10^-4 coulombs
the charge for the 10 microfarad capacitor is Q = 10*10^-6(50) = 5*10^-4 coulombs

and since they are in parallel, the Q net is the sum so you get Qsum = 6.5*10^-4 coulombs

but now i need to factor in the dielectric constant, kappa = 4 now for the 5 microfarad capacitor, correct? so C = kappa(Q/V) = 4(5*10^-6) =2*10^-5 farads

what changes when the positive sides are connected versus positive-negative? how does it affect the C = Q/V equation? does the magnitude of Q change?
 
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  • #6
Doc Al
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so the dielectric oil increases the capacitance and the charge increases too using C = Q/V and solving for Q
Careful! Inserting the dielectric increases the capacitance, but the charge can't change. It's stuck--no place to go! (Only when you put the two capacitors together can the charges move around.)
 
  • #7
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did you see my edits? i have some other questions relating the dielectric's effect on the capacitance equation..above..
 
  • #8
Doc Al
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so the dielectric oil increases the capacitance and the charge increases too using C = Q/V and solving for Q
Read my comment above.
the charge for the 5 microfarad capacitor is Q = C_1V = 5*10^-6(50) = 2.5*10^-4 coulombs
the charge for the 10 microfarad capacitor is Q = 10*10^-6(50) = 5*10^-4 coulombs
OK.
and since they are in parallel, the Q net is the sum so you get Qsum = 6.5*10^-4 coulombs
If you connected them + to +, then that would be the total charge.
but now i need to factor in the dielectric constant, kappa = 4 now for the 5 microfarad capacitor, correct? so C = kappa(Q/V) = 4(5*10^-6) =2*10^-5 farads
OK.
what changes when the positive sides are connected versus positive-negative? how does it affect the C = Q/V equation? does the magnitude of Q change?
It changes the total charge. You'll have two equations:
V = Q1/C1 =Q2/C2
Q1 + Q2 = Q total (which you know from the original set up)
 
  • #9
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i think i may have miscalulated my Qsum, i forgot to include the potential V in it, originaly i had:
C = kappa(Q/V) = 4(5*10^-6) =2*10^-5 farads

it should be: C = kappa(Q/V) = [4(5*10^-6)]/50 =4*10^-7 farads

and now that we concluded that if we connected + to +, that the total charge would be 6.5*10^-4 coulombs, we now have C and Q for the equation C = Q/V

now part a asks me to determine the electric potential if the two positives are connected, this is what i did:

C = Q/V ---> V = CQ where C = 4*10^-7 farads, Q = 6.5*10^-4 coulombs, so i got V = 1625 volts for part a

did i do part a correctly?

now i tried part b, where the - of one capacitor is connected to the + of the other, here's what i did:

since V = Q1/C1 = Q2/C2 and Q1 + Q2 = Q total

solving the Q eq, i got Q1 + Q2 = 6.5*10^-4, so Q1 = 6.5*10^-4 - Q2, and then using the other eq for V, i set up this equality
V = Q1/C1 = (6.5*10^-4 - Q2)/C1 where Q2 = 5*10^-4, C1 = 4*10^-7 <from above> and then solved for V

i got V = 375 volts for part b? did i do part b correctly?
 
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  • #10
Doc Al
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i think i may have miscalulated my Qsum, i forgot to include the potential V in it, originaly i had:
C = kappa(Q/V) = 4(5*10^-6) =2*10^-5 farads

it should be: C = kappa(Q/V) = [4(5*10^-6)]/50 =4*10^-7 farads
You were right the first time. Realize that here you're not really calculating C using Q/V: You don't have Q yet! All you're doing is multiplying the original capacitance by the dielectric constant, as explained here: http://hyperphysics.phy-astr.gsu.edu/hbase/electric/dielec.html#c2".

Revise what you've done accordingly and you'll be OK.
 
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  • #11
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so used what i originally had for capacitance with the dielectric (+ to +):
C = kappa(Q/V) = 4(5*10^-6) =2*10^-5 farads

C = Q/V ---> V = CQ where C = 2*10^-5 farads, Q = 6.5*10^-4 coulombs, so i got V = 32.5 volts for part a

part b (+ to - ):

since V = Q1/C1 = Q2/C2 and Q1 + Q2 = Q total

solving the Q eq, i got Q1 + Q2 = 6.5*10^-4, so Q1 = 6.5*10^-4 - Q2, and then using the other eq for V, i set up this equality:

V = Q1/C1 = (6.5*10^-4 - Q2)/C1 where Q2 = 5*10^-4, C1 = 2*10^-5 <from above> and then solved for V = 7.5 volts

are my new potentials for part a and b correct now?

thanks
 
  • #12
Doc Al
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You made an error in calculating the total charge, which I didn't catch before:
the charge for the 5 microfarad capacitor is Q = C_1V = 5*10^-6(50) = 2.5*10^-4 coulombs
the charge for the 10 microfarad capacitor is Q = 10*10^-6(50) = 5*10^-4 coulombs

and since they are in parallel, the Q net is the sum so you get Qsum = 6.5*10^-4 coulombs
Also: The total charge on the connected capacitor plates is different for parts a and b. In part a, you are adding two positive charges; for part b, you are adding a positive and a negative charge.
 
  • #13
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okay i guess i made a simple addition error, should Qsum = 7.5*10^-4 coulombs, not 6.5*10^-4
my new electric potential for part a is V = 37.5volts

as for part b, do i just change the sign of Q2, here is what i get now:
it use to be this:

" since V = Q1/C1 = Q2/C2 and Q1 + Q2 = Q total

solving the Q eq, i got Q1 + Q2 = 6.5*10^-4, so Q1 = 6.5*10^-4 - Q2, and then using the other eq for V, i set up this equality:

V = Q1/C1 = (6.5*10^-4 - Q2)/C1 where Q2 = 5*10^-4, C1 = 2*10^-5 <from above> and then solved for V = 7.5 volts "

now i did this:

since V = Q1/C1 = Q2/C2 and Q1 + Q2 = Q total

solving the Q eq, i got Q1 + Q2 = 7.5*10^-4, so Q1 = 7.5*10^-4 + Q2, and then using the other eq for V, i set up this equality:

V = Q1/C1 = (6.5*10^-4 + Q2)/C1 where Q2 = 5*10^-4, C1 = 2*10^-5 <from above> and then solved for

V = 62.5 volts

is that what i was supposed to do for part b?
 
  • #14
Doc Al
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okay i guess i made a simple addition error, should Qsum = 7.5*10^-4 coulombs, not 6.5*10^-4
my new electric potential for part a is V = 37.5volts
Show how you got this answer for V.

as for part b, do i just change the sign of Q2, here is what i get now:
it use to be this:

" since V = Q1/C1 = Q2/C2 and Q1 + Q2 = Q total
Good.

solving the Q eq, i got Q1 + Q2 = 6.5*10^-4
How did you get the 6.5?
 
  • #15
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where you saw this "solving the Q eq, i got Q1 + Q2 = 6.5*10^-4 " i was just showing how it was before i changed it, i got 6.5 by making a careless addition mistake of 2.5... + 5..., it was meant to be 7.5... which i corrected above

this is how i did part a:

so used what i originally had for capacitance with the dielectric (+ to +):
C = kappa(Q/V) = 4(5*10^-6) =2*10^-5 farads

C = Q/V ---> V = CQ where C = 2*10^-5 farads, Q = 7.5*10^-4 coulombs, so i got V = 37.5 volts for part a
 
  • #16
Doc Al
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this is how i did part a:

so used what i originally had for capacitance with the dielectric (+ to +):
C = kappa(Q/V) = 4(5*10^-6) =2*10^-5 farads

C = Q/V ---> V = CQ where C = 2*10^-5 farads, Q = 7.5*10^-4 coulombs, so i got V = 37.5 volts for part a
This is not correct. Here Q stands for the total charge on both capacitors, not the charge on one of them. So you can't use V = Q/C until you solve for the charge on that capacitor.

Parts a and b need to be solved in the same fashion. Set up your equations for total charge and voltage. Then you can solve for the charge on each capacitor, which will allow you to calculate the voltage.
 

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