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 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(1e[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/learningcenter/calculators/133wireresistancecalculatortable
$$C=K\varepsilon_0\frac{A}{d}=C=\frac{(8.85\times10^{12})\times1\times 0.0254469}{0.01}=0.000022531F=22.531μF$$
6V*(12.71828^{t/0.019Ω*0.0000225307uF})
time constants
6*(12.71828^{1}) = 6*(0.632) = 3.79 sec
6*(12.71828^{2}) = 6*(0.865) = 5.18 sec
6*(12.71828^{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
6V*(12.71828^{t/0.019Ω*0.0000225307uF})
time constants
6*(12.71828^{1}) = 6*(0.632) = 3.79 sec
6*(12.71828^{2}) = 6*(0.865) = 5.18 sec
6*(12.71828^{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