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 = i1 + i2, V1 = q1 *1/C1, V2 = q2 *1/C2, V = V1 = V2Taking the time derivative of the second and third equations, we can write:
dV1 /dt= i1 *1/C1, dV2/dt= i2 *1/C2Substituting into i = i1 + i2 and noting that V1' = V2' =V' :
i = V'C1 + V'C2 = V'*(C1 + C2)or, rewriting:
V' = i * 1/(C1 + C2)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 = C1 + C2Compare 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.