Capacitors are passive two terminals electronic components that store energy in form of electrical field. Practical capacitors contain at least two electrical conductors (plates) separated by a dielectric that is; insulator. The conductors can be thin “non-conducting”. Dielectric acts to increase the capacitance. That is; capacitor’s ability to store charge.

A dielectric can however be glass, ceramic, plastic film, air, paper, mica, etc. Unlike resistors, capacitors do not dissipate energy.

A capacitor stores energy in the form of an electrostatic field between its plates. Attaching a capacitor across a battery, will hence cause positive charge (᷿+Q) to collect on one plate and also negative charge (᷿-Q) on the other plate. Meanwhile, If a battery has been connected to a capacitor for a sufficient amount of time, no current can flow through the capacitor. However, if an alternating voltage is applied across the leads of the capacitor, a displacement current can flow.

The potential difference (V) across a capacitor is always directly proportional to the charge (Q) on either of its plates. The more the charge, the stronger the electric field between the plates, and thereby greater it is the potential difference.

V ∝ Q

The ratio between charge (Q) and potential difference (V) is therefore a constant for any capacitor. This is known as its capacitance (C).


Capacitors are used along with resistors in timing circuits because it takes time for a capacitor to fill with charge. They are used to smooth varying D.C supplies by acting as a reserviour of charge. Also, used in filter circuits because they easily pass alternating (changing) signals but block direct (constant) signals.


The Standard International (S.I) unit of capacitor’s capacitance is the “Farad” abbreviated “F” where;


i.e C=Q/V

The Farad is so large a unit for practical purposes. Therefore, it is usually replaced by either microfarad (µF) or nano farad (nF), and or pico farad (pF) whose values are;

1µF = 10-6F

1nF = 10-9F

1pF = 10-12F


Calculate the charge between the plates of a capacitor of capacitance of 10µF. Given that a potential difference (Pd) of 100V is applied across it.



Therefore; Q=V/C


=10Colombs (C)


A capacitor stored  charge of 50C across . If the capacitance of the capacitor was 47µF, calculate the potential difference (Pd) applied across it.



Therefore; V=CQ





Types of Capacitors / Polarized Capacitors / Non-Polarized Capacitors / Variable Capacitors / Capacitors in Series Calculations / Capacitors in Parallel Calculations

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