Energy in a capacitor, dielectrics and the electric dipole

Last week you computed how much charge a capacitor holds. This week you answer three questions: how much energy is in a capacitor, what changes when you put an insulator between the plates, and how an electric dipole behaves in a uniform field. The pattern that comes up most often on the exam is “a dielectric was added, what happened to this quantity?” The answer depends entirely on whether the battery is connected.

Golden rules:

1. $U = \tfrac12 C V^2 = \dfrac{Q^2}{2C} = \tfrac12 QV$. All three are the same; use the one in which a quantity stays constant.
2. A dielectric increases C by a factor κ: $C = \kappa C_0$.
3. If the battery is removed, Q is constant; if it is connected, V is constant. Everything else follows from this.
4. In a uniform field the net force on a dipole is zero, and the torque is $pE\sin\theta$.
5. The energy of a dipole is $U = -pE\cos\theta$: lowest when it is parallel to the field.

Energy stored in a capacitor

Charging a capacitor takes work. When the plates have charge $q$, the voltage is $q/C$. The work to move the next small charge $dq$ from the negative plate to the positive plate is $dW = (q/C)\,dq$. Add from zero to $Q$:

$U = \int_0^Q \frac{q}{C}\,dq = \frac{Q^2}{2C}$

Substituting $Q = C\Delta V$ gives the other two forms:

$U = \frac{Q^2}{2C} = \frac12 Q\,\Delta V = \frac12 C(\Delta V)^2$

Why half of $Q\Delta V$? Because while the charge was being moved, the voltage rose slowly from zero to $\Delta V$. The average voltage is $\Delta V/2$.

Example 1. A $20\ \mu\text{F}$ capacitor is charged to 50 V.

U = ½ C V² = ½ (20e-6)(50)² = ½ (20e-6)(2500) = 0.025 J = 25 mJ

Energy density

Where is the energy? In the electric field between the plates. For the parallel-plate capacitor, writing $C = \varepsilon_0A/d$ and $\Delta V = Ed$ gives $U = \tfrac12\varepsilon_0E^2(Ad)$. Since $Ad$ is the volume, the energy per unit volume is:

$u_E = \frac12\varepsilon_0E^2 \qquad (\text{J/m}^3)$

This formula was derived from the parallel-plate capacitor but is valid for any electric field: where there is a field there is energy.

Example 2. $A = 0.010\ \text{m}^2$, $d = 2.0\ \text{mm}$, $\Delta V = 400\ \text{V}$. Find the energy in two ways.

C = ε0 A/d = (8.85e-12)(0.010)/(2.0e-3) = 44.3 pF
U = ½ C V² = ½ (44.3e-12)(400)² = 3.54e-6 J
E = V/d = 400/2.0e-3 = 2.0e5 V/m
u = ½ ε0 E² = ½ (8.85e-12)(2.0e5)² = 0.177 J/m³
U = u · A d = 0.177 · (0.010 · 2.0e-3) = 3.54e-6 J   (the same)

PITFALL · The energy of a single capacitor in a series circuit. When finding the energy of one capacitor in series, don’t put the source voltage directly into $\tfrac12CV^2$. Use that capacitor’s own voltage (or the common $Q$): $U_i = Q^2/(2C_i)$.

Energy lost in charge sharing

When a charged capacitor that has been disconnected from the battery is connected to an empty capacitor, charge is conserved and energy is not. The final energy is always smaller than the initial; the difference goes as heat and electromagnetic radiation in the wires. Calculation: first $V_{final} = Q_0/(C_1+C_2)$, then $U_f = \tfrac12(C_1+C_2)V_{final}^2$.

Dielectrics

A dielectric is an insulating material placed between the plates: paper, glass, plastic, oil. If it completely fills the gap, the capacitance increases by the dielectric constant κ:

$C = \kappa C_0, \qquad C = \kappa\frac{\varepsilon_0 A}{d}$

κ is dimensionless and always greater than or equal to 1. It is exactly 1 for vacuum, 1.00059 for air (1 in practice), about 3.7 for paper, about 80 for water.

A dielectric has three benefits:

1. It increases the capacitance.
2. It keeps the plates mechanically apart, making a very small $d$ possible.
3. Its dielectric strength is higher than that of air. Dielectric strength is the largest field the material can withstand without breaking down (becoming conducting). The largest voltage that can be applied to the capacitor is $\Delta V_{max} = E_{max}d$. For example, for a sheet $0.10\ \text{mm}$ thick with a strength of $16\times10^{6}\ \text{V/m}$, $\Delta V_{max} = 1.6\ \text{kV}$. For air this value is around $3\times10^{6}$ V/m.

Battery connected or removed?

Memorize this table; half the exam question is this.

Example 3. $C_0 = 50\ \text{pF}$, $\Delta V_0 = 100\ \text{V}$, then a dielectric with $\kappa = 4$ fills the gap completely. Compare the two cases.

Initial:  Q0 = C0 V0 = 5.0 nC,  U0 = ½ C0 V0² = 0.25 μJ
C = 4 · 50 = 200 pF (in both cases)

Removed (Q constant):
  V = Q0/C = 5.0 nC / 200 pF = 25 V
  U = Q0²/(2C) = 0.0625 μJ = U0/4

Connected (V constant):
  Q = C V0 = 200 pF · 100 V = 20 nC
  U = ½ C V0² = 1.0 μJ = 4 U0

TIP · Why did the energy decrease? In the removed case the dielectric is pulled in between the plates. The field does work on the slab, and this work comes from the stored energy. In the connected case the battery pushes extra charge and supplies the extra energy.

Partly filled capacitor

If the dielectric does not fill the gap completely, divide the capacitor into parts:

• The slab is parallel to the plates and fills part of the thickness: the same charge passes through both layers, two capacitors in series.
• The slab fills part of the area at full thickness: the two regions are at the same voltage, two capacitors in parallel.

Write the capacitance of each part as $\kappa\varepsilon_0A_i/d_i$, then apply last week’s series/parallel rules.

Electric dipole in a uniform field

An electric dipole is a charge $+q$ and a charge $-q$ separated by a distance $2a$. The dipole moment:

$\vec p = q\,(2a)\,\hat u, \qquad p = 2aq \quad (\text{C·m})$

Its direction is from the negative charge to the positive charge.

Force and torque

In a uniform field $\vec E$, the force $q\vec E$ acts on $+q$ and $-q\vec E$ on $-q$. The total force is zero, so the dipole has no translational motion. But because the two forces act at different points, they form a couple. About the center, the moment arm of each force is $a\sin\theta$:

$\tau = 2(qE)(a\sin\theta) = pE\sin\theta, \qquad \vec\tau = \vec p\times\vec E$

The torque tries to turn the dipole toward the direction of the field. At $\theta = 0$ the torque is zero and the equilibrium is stable; at $\theta = 180^\circ$ the torque is again zero but the equilibrium is unstable.

Potential energy

The work done to rotate the dipole is stored as potential energy. The reference is chosen as $U = 0$ at $\theta = 90^\circ$:

$U = -\vec p\cdot\vec E = -pE\cos\theta$

The work done by an external agent to rotate from $\theta_1$ to $\theta_2$ is $W = U(\theta_2) - U(\theta_1)$. From 0 to 180°: $W = 2pE$.

Example 4. Charges $\pm2.0\ \text{nC}$ are $3.0\ \text{mm}$ apart, $E = 5.0\times10^{4}\ \text{N/C}$, and the dipole makes an angle of $30^\circ$ with the field.

p = q · 2a = (2.0e-9)(3.0e-3) = 6.0e-12 C·m
τ = pE sinθ = (6.0e-12)(5.0e4)(0.500) = 1.5e-7 N·m
U = −pE cosθ = −(6.0e-12)(5.0e4)(0.866) = −2.6e-7 J
Work to rotate 0° → 180°: 2pE = 6.0e-7 J

Molecules with a permanent dipole moment, such as the water molecule, are called polar molecules. A microwave oven applies this torque to water molecules with a field that keeps changing direction. The atomic explanation of dielectrics also rests on exactly this dipole idea; we will see it in week 10.

Common mistakes