# Parallel Plate Capacitance and Charge Time

#### HelloCthulhu

Homework Statement
A 6V battery, 0.64mm diameter/14in length single/solid copper wire alligator clips how long and an 18cm disc parallel plate air capacitor are used in a circuit. What is the capacitance? How long will it take to charge the capacitor to 100% of its capacity?
Homework Equations
A=πr2

capacitance
$$C=K\varepsilon_0\frac{A}{d}$$
Area of 18cm disc electrode = 0.0254469m2
d of separation = 0.01m
dielectric constant = 1
ε0 = 8.85×10−12 F/m−1

voltage respect to time
V(t) = VB(1-e[SUP]-t/RC[/SUP])
V(t) = voltage in respect to time
VB = battery voltage
-e (numerical constant) = 2.71828
-t = voltage in respect to time in sec
R = resistance of the wire
C = capacitance
RC = time constant = τ

resistance for the 14in long/ 0.64mm diameter wire found here:
https://www.cirris.com/learning-center/calculators/133-wire-resistance-calculator-table
$$C=K\varepsilon_0\frac{A}{d}=C=\frac{(8.85\times10^{-12})\times1\times 0.0254469}{0.01}=0.000022531F=22.531μF$$

6V*(1-2.71828-t/0.019Ω*0.0000225307uF)

time constants
6*(1-2.71828-1) = 6*(0.632) = 3.79 sec
6*(1-2.71828-2) = 6*(0.865) = 5.18 sec
6*(1-2.71828-3) = 6*(0.950) = 5.70 sec
6*(0.982) = 5.89 sec
6*(0.993) = 5.95 sec

capacitance = 22.531μF
100% charge time = 6 sec

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#### Orodruin

Staff Emeritus
Homework Helper
Gold Member
2018 Award
0.993 is not equal to one. You also did not solve for t so your equations are dimensionally inconsistent.

#### HelloCthulhu

0.993 is not equal to one. You also did not solve for t so your equations are dimensionally inconsistent.
I apologize this is my first time doing this kind of equation. I used this video to learn about time constants:

In this equation I still have the -t value over the RC value:
6V*(1-2.71828-t/0.019Ω*0.00002253)
After the RC value was solved, I should've written these equations as:
6V*(1-2.71828-4.280833e-7/4.280833e-7)
6V*(1-2.71828-1)
6V*(0.632) = 3.79 sec

#### Orodruin

Staff Emeritus
Homework Helper
Gold Member
2018 Award
Again, you are being dimensionally inconsistent. 6 V multiplied by a dimensionless number still has units of volts. What you have computed is the capacitor charge after a time t=RC. This is not what the question asked for, the question asked when the capacitor would be fully charged, ie, when (for what t) does the capacitor charge become 6 V?

#### HelloCthulhu

Ok, I was definitely confused earlier. I found a more detailed tutorial on this, so I hope I have a better understanding of time constants and charging time now:

1t = 0.019Ω*0.00002253uF = 4.280833e-7 secs
5t = 5X4.280833e-7 = 0.00000214041 secs

#### gneill

Mentor
You might want to check your exponent calculations for the capacitance calculation. I can't imagine how an 18 cm disc capacitor separated by 10 mm of air can give such a large capacitance (the numbers of the mantissa look good, but I think the power of 10 you've ascribed to the exponent is incorrect).

#### HelloCthulhu

You might want to check your exponent calculations for the capacitance calculation. I can't imagine how an 18 cm disc capacitor separated by 10 mm of air can give such a large capacitance (the numbers of the mantissa look good, but I think the power of 10 you've ascribed to the exponent is incorrect).
You're absolutely right. I submitted 22.531μF as the solution to the capacitance equation. But the capacitance is 2.25307x10-11F (0.0000225307μF).

$$C=K\varepsilon_0\frac{A}{d}=C=\frac{(8.85\times10^{-12})\times1\times0.0254469}{0.01}=2.25307\times10^{-11}F=0.0000225307μF$$

I used this online calculator to check the work. Hope I got it right this time.

#### gneill

Mentor
Yup. So think pF rather than $\mu F$ for the capacitance value.

Your idea of suggesting that a period of 5 time constants will settle the matter is good in "engineering" terms. It is true that the engineering rule of thumb is that everything of interest in a circuit has occurred after a period of five time constants. However, be aware that theoretically, a charging (or discharging) capacitor will never reach it's final charge in a finite amount of time. In other words, this may be a trick question. You may want to review the original question as presented to you to be sure that it is asking for an "engineering" solution or a theoretical one.

"Parallel Plate Capacitance and Charge Time"

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