Capacitor Discharge problem

[haxxorboi]

Homework Statement

A capacitor is charged to 1 coulomb; the capacitance is 8.00×10-5 farads. Then a switch is closed which puts the capacitor in a closed circuit with a resistor; the resistance is 8.60×10¹ ohms. Calculate the charge on the capacitor after 4.00 milliseconds. (1 ms = 0.001 s)

Homework Equations

I=V/R
C=Q/V

The Attempt at a Solution

C=Q/V=8E-5 F=1/V V=12500 V

I=V/R=12500/86=145.348 A

145.348 C/S = .145348 C/mS

.145348*4=.5814 C transferred in that 4 milliseconds

1-.5814=.4186 C left

So the answer would be .4186 C I think but I'm getting an "Incorrect Answer" apparently

 

[LowlyPion]

Homework Statement

A capacitor is charged to 1 coulomb; the capacitance is 8.00×10-5 farads. Then a switch is closed which puts the capacitor in a closed circuit with a resistor; the resistance is 8.60×10¹ ohms. Calculate the charge on the capacitor after 4.00 milliseconds. (1 ms = 0.001 s)

Homework Equations

I=V/R
C=Q/V

The Attempt at a Solution

C=Q/V=8E-5 F=1/V V=12500 V

I=V/R=12500/86=145.348 A

145.348 C/S = .145348 C/mS

.145348*4=.5814 C transferred in that 4 milliseconds

1-.5814=.4186 C left

So the answer would be .4186 C I think but I'm getting an "Incorrect Answer" apparently

This is a time varying current flow. You will want to consider the Charge as a function of time.

$$Q_{(t)} = Q_o*e^{\frac{-t}{RC}}$$

 

[haxxorboi]

This is a time varying current flow. You will want to consider the Charge as a function of time.

$$Q_{(t)} = Q_o*e^{\frac{-t}{RC}}$$

I've never seen that equation before and I can't find it in our chapter.

Let me see if I understand this equation...

Final charge=1 Coulomb * 1.602E-19^{-.004/(86*8E-5)}

Does that seem right or am I misinterpreting part of the equation.

Thanks

 

[Phrak]

"e" is the natural base = 2.71828..., rather than the charge of the electron, in this case. Just write it as "e", and use the exponental function on your calculator.

 

[LowlyPion]

"e" is the natural base = 2.71828..., rather than the charge of the electron, in this case. Just write it as "e", and use the exponental function on your calculator.

This is correct. Not e the charge of an electron. e the natural log

It's typically used to express the decay in RC networks.

 

[haxxorboi]

SWEET! The answer ended up being .5591 C which was correct with the system.

Thanks so much for the help guys, I feel stupid with not recognizing which "e" that was seeing as this is an exponential decay. I'm not sure how we were supposed to know how to do this problem though as I can't even find this equation in our course pack I have no idea how my Prof expected our Phy101 class to know this without going for help.

Thanks again guys for the help!

 

[Wizardofwaz]

I have the same problem but with these numbers
A capacitor is charged to 1 coulomb; the capacitance is 6.00×10-5 farads. Then a switch is closed which puts the capacitor in a closed circuit with a resistor; the resistance is 9.20×101 ohms. Calculate the charge on the capacitor after 4.00 milliseconds. (1 ms = 0.001 s)

what kind of formula do I use to solve this