##
Capacitors in parallel

What is the capacitance of this parallel combination of two identical capacitors?

Let's first draw this as a simple diagram:

Note that I have indicated a current coming into the system, and I labeled
the distinct currents through the capacitors. From Kirchoff's Laws and
the constitutive laws, we know:
i = i_{1 }+ i_{2}, V_{1} = q_{1
}*1/C_{1}, V_{2} = q_{2 }*1/C_{2},
V = V_{1} = V_{2}

Taking the time derivative of the second and third equations, we can write:
dV_{1} /dt= i_{1 }*1/C_{1}, dV_{2}/dt=
i_{2 }*1/C_{2}

Substituting into i = i_{1 }+ i_{2} and noting that V_{1}'
= V_{2}' =V' :
i = V'C_{1} + V'C_{2} = V'*(C_{1} +
C_{2})

or, rewriting:
V' = i * 1/(C_{1} + C_{2})

Now, I have a relationship between the voltage across both capacitors and
the total current running into the system of the form, V' = i * 1/C. Thus,the
net capacitance of two capacitors in parallel is :
C = C_{1} + C_{2}

Compare this result with that for two
resistors in series or two resistors in parallel. Note that since the
constitutive law for a resistor is V=iR, while for a capacitor is V=q*1/C,
the formula for the effective capacitance of two capacitors in parallel
looks like that for two resistors in series. Similarly, you can show that
two capacitors in series will have an effective capacitance of a form like
that for two resistors in parallel!
Finally, the final answer for the capacitance of the picture above is
simplified since the two capacitors are identical. Thus, the effective
capacitance will be C = 2 * 510mF = 1020mF.