Capacitor And Capacitance - Practice Questions & MCQ

A capacitor is a passive two-terminal electrical component used to store energy electrostatically in an electric field. The forms of practical capacitors vary widely, but all contain at least two electrical conductors (plates) separated by a dielectric  (i.e., insulator). The conductors can be thin films of metal, aluminum foil or disks, etc. The 'nonconducting' dielectric acts to increase the capacitor's charge capacity. A dielectric can be glass, ceramic, plastic film, air, paper, mica, etc. Capacitors are widely used as parts of electrical circuits in many common electrical devices. Unlike a resistor, a capacitor does not dissipate energy. Instead, a capacitor stores energy in the form of an electrostatic field between its plates.

Capacitance: Capacitance is the ability of a capacitor to hold an electrical charge. Capacitors are components in an electrical circuit that can store a charge and are considered one of the three fundamental electronic components along with inductors and resistors.

For an isolated capacitor, capacitance is defined as the ratio of the charge to the potential of a capacitor.

$
C=\frac{Q}{V}
$

As Q increases V also increases because potential depends on the charge.

$
\begin{gathered}
Q \propto V \\
Q=C V
\end{gathered}
$

Therefore Capacitance is a constant quantity.
Unit of Capacitance-

$
\text { S.Iunit }-\frac{C}{V}=\operatorname{farad}(f)
$

Smaller S.I unit is $m f, \mu f, n f$ and $P f$
C.G.S - Stat farad and $1 F=9 \times 10^{11}$ statfarad

Dimension $=M^{-1} L^{-2} T^4 A^2$
- If $\mathrm{V}=1 \mathrm{~V}, \mathrm{C}=\mathrm{Q}$. Hence, we define the capacitance of an conductor as the charge required to rise the potential of the conductor by 1 V . In the SI system, the unit of capacitance is farad.

CAPACITANCE OF A SPHERICAL CONDUCTOR OR CAPACITOR:
A single conductor can also act as a capacitor. For this, let a charge q be given to a spherical conductor of radius R , then potential on it is

$
V=\frac{1}{4 \pi \varepsilon_0} \frac{q}{R}
$

The other conductor is supposed to be at infinity, whose potential will be taken as zero. So the potential difference between the sphere and the conductor at infinity becomes $\mathrm{V}-0=\mathrm{V}$.

$
C=\frac{q}{V}=4 \pi \varepsilon_0 R
$

Thus, the capacitance of a spherical conductor is $C=4 \pi \varepsilon_0 R$, so we can C depends on the medium and dimension of conductor.