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Capacitance Formula

Capacitance is all about how much charge can any conductor hold. It is the ratio of the charge flowing across the conductor to the potential applied across it. The conductors used for holding charges are called as capacitors.

Capacitance is the ability of a body to store an electrical charge. Any object that can be electrically charged exhibits capacitance. A common form of energy storage device is a parallel-plate capacitor. Capacitance is typified by a parallel plate arrangement and is defined in terms of charge storage. When a capacitor is fully charged there is a potential difference, between its plates, and the larger the area of the plates and/or the smaller the distance between them, the greater will be the charge that the capacitor can hold and the greater will be its Capacitance.

Capacitance Formula is given by
Where,
Q is the the charge on the conductor,
V is the potential applied across the conductor and
C is proportionality constant called as capacitance.

If capacitors are connected in series, the capacitance formula is given by

If capacitors are connected in parallel, the capacitance formula is given by
Where C1,C2,C3.......Cn are the capacitors. Capacitance is expressed in Farads.

Capacitance formula is used in finding the capacitance of any number of capacitors in the given circuit.

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Capacitance Problems

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Problems based on capacitance are mentioned below:

Solved Examples

Question 1: Find the capacitance of the capacitor if 5 coulomb of charge is flowing and potential applied is 2 V.
Solution:
 
Given: Charge Q = 5 C,
          Voltage applied V = 2 V
The Capacitance is given by C = $\frac{Q}{V}$
                                             = $\frac{5 C}{2 V}$
                                             = 2.5 F.
 

Question 2: Find capacitance if capacitors 6 F and 5 F are connected
(i) In series
(ii) In parallel.
Solution:
 
The capacitance in series is given by Cs = $\frac{1}{C_{1}}$ + $\frac{1}{C_{2}}$
                                                           = $\frac{C_{1} + C_{2}}{C_{1} C_{2}}$
                                                           = $\frac{6 + 5}{30}$
                                                           = 0.367 F.
The capacitance in parallel is given by Cp = C1 + C2
                                                            = 6 + 5
                                                            = 11 F.
 

More topics in Capacitance Formula
Electrical Formulas Capacitive Reactance Formula
Time Constant Formula
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