Capacitance of a Parallel Plate Capacitor Without and With Dielectric Medium Between the Plates

(a) Capacitance of a parallel plate capacitor without a dielectric

Set-up and diagram

• A parallel plate capacitor consists of two thin conducting plates, each of area $A$, held parallel to each other at a distance $d$ apart. One plate is isolated and charged, the other is earthed. Draw the two plates, mark $+Q$ on the isolated plate and $-Q$ on the inner face of the earthed plate, and show the uniform field lines running from the positive to the negative plate.

Derivation

• When a charge $+Q$ is given to the isolated plate, a charge $-Q$ is induced on the inner face of the earthed plate and $+Q$ on its outer face; being free and earthed, that $+Q$ flows to earth.
• In the outer regions the fields of the two charged plates are oppositely directed and cancel:

$$E = \dfrac{\sigma}{2\varepsilon_0}-\dfrac{\sigma}{2\varepsilon_0} = 0$$

• In the region between the plates the two fields add:

$$E = \dfrac{\sigma}{2\varepsilon_0}+\dfrac{\sigma}{2\varepsilon_0} = \dfrac{\sigma}{\varepsilon_0} = \dfrac{Q}{A\varepsilon_0}\qquad \ldots (1)$$

• directed from the positive to the negative plate.
• If $V$ is the potential difference between the plates, the uniform field is

$$E = \dfrac{V}{d}\qquad\text{or}\qquad V = Ed\qquad \ldots (2)$$

• Substituting (1) in (2),

$$V = \dfrac{Qd}{A\varepsilon_0}$$

• Hence the capacitance is

$$\boxed{\,C = \dfrac{Q}{V} = \dfrac{Q}{\dfrac{Qd}{A\varepsilon_0}} = \dfrac{\varepsilon_0 A}{d}\,}$$

• Capacitance rises with plate area and falls with plate separation. For circular plates of radius $r$, put $A = \pi r^2$.

(b) Capacitance with a dielectric slab between the plates

Set-up

• Let $E_0$ be the field between the plates before the slab is introduced, so $V_0 = E_0 d$, where $E_0 = \dfrac{\sigma}{\varepsilon_0} = \dfrac{Q}{A\varepsilon_0}$ and $\sigma$ is the surface charge density on the plates. Without the slab, $C_0 = \dfrac{\varepsilon_0 A}{d}$.
• Introduce a dielectric slab of thickness $t\ (t<d)$. The field $E_0$ polarises the dielectric, inducing a charge $-Q_p$ on its left face and $+Q_p$ on its right face. Draw the two plates, the slab of thickness $t$ occupying part of the gap, and the induced surface charges on its faces opposing the plate charges.

Derivation

• The induced charges set up a field $E_p$ inside the dielectric, opposite to $E_0$:

$$E_p = \dfrac{\sigma_p}{\varepsilon_0} = \dfrac{Q_p}{A\varepsilon_0},\qquad \sigma_p = \dfrac{Q_p}{A}$$

• The net field inside the dielectric reduces to

$$E = E_0-E_p = \dfrac{E_0}{k}$$

• where $k$ is the dielectric constant. Hence $E = \dfrac{Q}{Ak\varepsilon_0}$, or equivalently $Q = Ak\varepsilon_0 E$.
• The reduced field $E$ exists over the distance $t$ and the original field $E_0$ over the remaining distance $(d-t)$, so the potential difference between the plates is

$$V = E_0(d-t) + Et = E_0(d-t)+\dfrac{E_0}{k}t = E_0\left(d-t+\dfrac{t}{k}\right)$$

$$V = \dfrac{Q}{A\varepsilon_0}\left(d-t+\dfrac{t}{k}\right)$$

• Therefore

$$\boxed{\,C = \dfrac{Q}{V} = \dfrac{\varepsilon_0 A}{d-t+\dfrac{t}{k}}\,}$$

Special cases

• Dielectric fills the entire space, $t = d$:

$$C = \dfrac{k\varepsilon_0 A}{d} = kC_0\qquad\Rightarrow\qquad k = \dfrac{C}{C_0}$$

• so the capacitance of a parallel plate capacitor increases $k$ times.
• $n$ dielectric slabs of thicknesses $t_1, t_2,\ldots,t_n$ stacked between the plates — equivalent to $n$ capacitors in series:

$$C = \dfrac{\varepsilon_0 A}{\dfrac{t_1}{k_1}+\dfrac{t_2}{k_2}+\cdots+\dfrac{t_n}{k_n}}$$

• $n$ dielectrics placed side by side, with plate areas $A_1, A_2,\ldots,A_n$ and separation $d$ — equivalent to $n$ capacitors in parallel:

$$C = \dfrac{\varepsilon_0}{d}\left(A_1k_1+A_2k_2+\cdots+A_nk_n\right)$$

• and if all the areas are equal to $\dfrac{A}{n}$, then $C = \dfrac{\varepsilon_0 A}{nd}\left(k_1+k_2+\cdots+k_n\right)$.
• A conducting slab of thickness $t$ inserted, for which $k=\infty$:

$$C = \left(\dfrac{d}{d-t}\right)C_0\qquad\Rightarrow\qquad C > C_0$$

• so the capacitance increases by the factor $\dfrac{d}{d-t}$, whatever the position of the slab. This is the reason a metal plate introduced between the plates of a charged capacitor increases its capacitance.
• If there are $n$ parallel plates, they form $(n-1)$ capacitors, so $C = \dfrac{(n-1)\varepsilon_0 A}{d}$.
• Spherical capacitor of concentric shells of radii $a$ (inner) and $b$ (outer): $C = 4\pi\varepsilon_0\left(\dfrac{ab}{b-a}\right)$; increasing the difference of the radii decreases the capacitance.
• Cylindrical capacitor of coaxial shells of radii $a$ and $b$ and length $l$: $C = \dfrac{2\pi\varepsilon_0 l}{\log_e\left(\dfrac{b}{a}\right)}$.

What changes when a dielectric is inserted

• Reverse the last column for the standard question in which the battery is first disconnected and the dielectric is then removed: capacitance decreases, charge stays constant, and potential difference, field and stored energy each increase $k$ times.

8.11 Displacement current (the section that follows)

• In the conducting part of a circuit, free electrons carry the current. Between the plates of a capacitor there are no free electrons in the dielectric, yet the circuit carries current while the capacitor charges.
• As current passes through the leads, the field between the plates increases and polarises the dielectric, so there is a current in the dielectric due to the movement of the bound charges. This is called displacement current or charge-separation current.
• From $q = Ak\varepsilon_0 E$, differentiating,

$$i_c = \dfrac{dq}{dt} = Ak\varepsilon_0\dfrac{dE}{dt}$$

• so for a fixed area, $i_c\propto\dfrac{dE}{dt}$: the rate of change of the field across the capacitor is directly proportional to the current in the conducting part of the circuit. The right-hand side has the dimensions of current and is the displacement current $i_d$, equivalent to $\dfrac{dq}{dt}$ in the conducting part.
• With no dielectric between the plates ($k=1$ for air or vacuum), $i_d = A\varepsilon_0\dfrac{dE}{dt}$.
• Generalisation: displacement currents are not confined to the space between capacitor plates. A displacement current exists at any point in space where a time-varying electric field exists, i.e. wherever $\dfrac{dE}{dt}\neq 0$.