Capacitance

What is Capacitance?

Capacitance is a measure of a capacitor's ability to store charge on its plates, in other words, its storage capacity.
The higher the capacitance of a capacitor, the greater is the amount of charge stored on the plates for the same applied voltage.
When a voltage source v is connected to the capacitor, as in Fig. 1(b), the source deposits a positive charge q on one plate and a negative charge -q on the other. The capacitor is said to store the electric charge.
A typical capacitor
(a)
A typical capacitor
(b)
Fig. 1: (a) A typical capacitor. (b) A capacitor with applied voltage v.
The amount of charge stored, represented by q, is directly proportional to the applied voltage v so that $$ \bbox[10px,border:1px solid grey]{q = Cv } \tag{1}$$ where C, the constant of proportionality, is known as the capacitance of the capacitor. The unit of capacitance is the farad (F), in honor of the English physicist Michael Faraday (1791-1867).
Capacitance is the ratio of the charge on one plate of a capacitor to the voltage difference between the two plates, measured in farads (F)
In particular, a capacitor has a capacitance of 1 F if 1 C of charge ( $6.242 \times 10^{18}$ electrons) is deposited on the plates by a potential difference of 1 V across its plates.
The farad, however, is generally too large a measure of capacitance for most practical applications, so the microfarad ( $10^{-6}$) or picofarad ($10^{-12}$) are more commonly encountered. Although the capacitance C of a capacitor is the ratio of the charge q per plate to the applied voltage v, it does not depend only on q or v. It also depends on the physical dimensions of the capacitor. For example, for the parallel-plate capacitor shown in Fig. 1, the capacitance is given by $$ \bbox[10px,border:1px solid grey]{C = {\epsilon A \over d}} \tag{2}$$ where A is the surface area of each plate, d is the distance between the plates, and $\epsilon$ is the permittivity of the dielectric material between the plates. Although Eq. (1) applies to only parallel-plate capacitors, we may infer from it that, in general, three factors determine the value of the capacitance: Capacitor voltage rating and capacitance are typically inversely rated due to the relationships in Eqs. (1) and (2) . Arcing occurs if d is small and V is high.
  • The surface area of the plates - the larger the area, the greater the capacitance.
  • The spacing between the plates - the smaller the spacing, the greater the capacitance.
  • The permittivity of the material - the higher the permittivity, the greater the capacitance.
For a vacuum, the value of $\epsilon$ (denoted by $\epsilon_o$) is $8.85 \times10^{-12}$ F/m. The ratio of the permittivity of any dielectric to that of a vacuum is called the relative permittivity, $\epsilon_r$. It simply compares the permittivity of the dielectric to that of air. In equation form, $$\bbox[10px,border:1px solid grey]{\epsilon_r = {\epsilon \over \epsilon_0} }\tag{3} $$ The value of e for any material, therefore, is $$\epsilon = \epsilon_r \epsilon_0 $$
Note that $\epsilon_r$ the relative permittivity, or dielectric constant, is a dimensionless quantity
To obtain the current-voltage relationship of the capacitor, we take the derivative of both sides of Eq. (1). Since $$ i = {dq \over dt } \tag{4}$$ differentiating both sides of Eq. (1) gives $$ {dq \over dt} = C {dv \over dt}$$ $$\bbox[10px,border:1px solid grey]{ i = C {dv \over dt}} \tag{5}$$ This is the current-voltage relationship for a capacitor, assuming the positive sign convention. The relationship is illustrated in Fig. 2 for a capacitor whose capacitance is independent of voltage. Capacitors that satisfy Eq. (5) are said to be linear. For a nonlinear capacitor, the plot of the current-voltage relationship is not a straight line. Although some capacitors are nonlinear, most are linear.
The voltage-current relation of the capacitor can be obtained by integrating both sides of Eq. (5). We get $$ v = {1 \over C} \int_{-\infty}^t i \, dt $$ or $$ \bbox[10px,border:1px solid grey]{v = {1 \over C} \int_{-\infty}^t i \, dt + v(t_0)} \tag{6} $$ where v(t=0) = q(t=0)/C is the voltage across the capacitor at time t=0. Equation (6) shows that capacitor voltage depends on the past history of the capacitor current. Hence, the capacitor has memory property that is often exploited. The instantaneous power delivered to the capacitor is $$p = vi = v (C{dv \over dt})$$ The energy stored in the capacitor is therefore $$ \begin{split} w &=\int_{-\infty}^t p \, dt =\int_{-\infty}^t v (C{dv \over dt}) \, dt \\ & = C \int_{-\infty}^t v \, dv = {1 \over 2} C v^2 \Big|_{\infty}^t \end{split}$$ We note that v(t=0) = 0, because the capacitor was uncharged at t =0. Thus, $$\bbox[10px,border:1px solid grey]{w = \tfrac{1}{2} C v^2} \tag{7}$$
Capacitance Related Questions