Electrostatic Potential and Capacitance

Written and reviewed by Akanksha Tiwari, Physics & Chemistry faculty at UniDrill (M.Sc. Physics, Banaras Hindu University · 8 years teaching Class XI–XII science). Based on NCERT Physics Class XII, Chapter 2.

Why can a bird sit on a high voltage wire?

A power line may be at thousands of volts, yet a bird perching on it is unharmed. The wire is at a high potential, but both of the bird's feet touch the same wire, so they are at the same potential. No potential difference across its body means no current through it. The danger comes from differences of potential, not from the potential itself. This is the central idea of the chapter: electric potential is the electrical equivalent of height. Nothing flows between two points at the same height.

From potential energy to potential

The electrostatic force is conservative. The work done in moving a charge between two points depends only on the points, not on the path, exactly like gravity. That lets us define a potential energy and, from it, a quantity that belongs to the field alone.

The electrostatic potential V at a point is the work done by an external agent in bringing a unit positive test charge from infinity to that point, without any acceleration. Its unit is the volt, $1\ \text{V} = 1\ \text{J C}^{-1}$. Only differences of potential have physical meaning: the potential at infinity is simply chosen to be zero. If a charge q is moved through a potential difference V, the work done is $W = qV$.

An everyday unit follows. One electron volt is the energy gained by an electron moving through 1 V: $1\ \text{eV} = 1.6 \times 10^{-19}\ \text{J}$. It is a unit of energy, not of voltage. An electron accelerated from rest through 100 V gains 100 eV, so $1/2\,mv^2 = 100 \times 1.6 \times 10^{-19}\ \text{J}$, giving a speed $v = \sqrt{2eV/m} \approx 5.9 \times 10^{6}\ \text{m s}^{-1}$.

Potential of a point charge, and why it is easier than the field

For a point charge Q the potential at distance r is

$V = kQ/r$, with $k = 9 \times 10^{9}\ \text{N m}^2\ \text{C}^{-2}$

It is positive for a positive Q and negative for a negative Q. It falls as $1/r$, whereas the field falls as $1/r^2$. The two are related by $E = kQ/r^2 = V/r$ for a point charge.

Here is the big advantage. Potential is a scalar. For several charges you add the potentials of each charge algebraically, with their signs. You never have to resolve components. The field of the same charges would need vector addition.

Worked example: work done by bringing in a charge. A charge of 4 × 10⁻⁷ C is fixed at a point. How much work is done in bringing a charge of 2 × 10⁻⁹ C from infinity to a point 9 cm from it?

The common wrong step: putting $r = 9$ in the formula instead of 0.09 m. This gives a result 100 times too large. The path does not matter, because the force is conservative.

Worked example: where is the potential zero? A charge +q and a charge −2q are 6 cm apart. Find the points on the line through them where the potential is zero.

The common wrong step: assuming $V = 0$ at the midpoint. That would need equal charges of opposite sign. Here the larger magnitude of −2q means the zero lies nearer to +q.

Potential of a short dipole. For a dipole of moment p at distance r (much larger than the separation), at an angle θ from the dipole axis,

$$V = \dfrac{kp\cos\theta}{r^2}$$

It falls as $1/r^2$, faster than a single charge, because the positive and negative contributions partly cancel. On the axis ($\theta = 0$ or $\pi$) the magnitude is largest. On the equatorial plane ($\theta = 90^\circ$), $V = 0$, even though the field there is not zero.

Pause and predict: four charges +q, −q, +q, −q are placed at the corners of a square, in that order around it. What are V and E at the centre? What if the order is +q, +q, −q, −q?

Check your answer: in both cases $V = 0$, because the distances are equal and the charges sum to zero. For the alternating order (+, −, +, −), E is also zero, since opposite corners are equal charges and their fields cancel. For (+, +, −, −), E is not zero: the fields of the adjacent like charges point the same way and add. So $V = 0$ never guarantees $E = 0$.

Charged spherical shell. For a thin shell of charge q and radius R, outside it ($r \ge R$) the potential is $kq/r$, exactly as if all the charge were at the centre. Inside ($r \le R$) the field is zero, so no work is done in moving a charge about, and the potential is the same everywhere inside: $V = kq/R$, equal to the surface value, and not zero.

Equipotential surfaces and the link with the field

An equipotential surface has the same potential at every point. Three properties follow from that definition.

The field is always perpendicular to the surface, because a tangential component of E would do work along the surface and change V. No work is done moving a charge along it. And two equipotential surfaces never intersect, because the point of intersection would have two different values of V.

The field and the potential are connected by

$$E = -\dfrac{dV}{dl}$$

The field points in the direction in which V falls fastest, and its size is the rate of fall. In a uniform field the potential difference over a distance d along the field is simply $V = Ed$. So E is also measured in volts per metre ($1\ \text{V m}^{-1} = 1\ \text{N C}^{-1}$). Where equipotential surfaces are close together, the field is strong.

Potential energy of charges

To assemble two charges q₁ and q₂ a distance r apart, an external agent must do work

$$U = \dfrac{kq_1q_2}{r}$$

It is positive for like charges (work is done against repulsion) and negative for unlike charges (they attract). For three charges, add the energy of every distinct pair, so there are three terms. A charge q placed in an external potential V(r) has potential energy $U = qV$.

Worked example: energy of a group of charges. (a) Charges +6 µC and −2 µC are 0.2 m apart. Find their potential energy and the work needed to separate them completely. (b) Three charges of +1 µC sit at the corners of an equilateral triangle of side 10 cm. Find the energy of the system.

The common wrong step: counting six terms in (b), once for each ordered pair. Each pair of charges shares one interaction energy, so count each pair once.

Dipole in a uniform field. A dipole in a uniform field E has potential energy $U = -pE\cos\theta = -p \cdot E$, with $U$ taken as zero at $\theta = 90^\circ$. It is lowest ($-pE$) when the dipole is aligned with the field, a stable equilibrium. It is highest ($+pE$) when opposite to the field, an unstable equilibrium. The work needed to turn a dipole from $\theta_1$ to $\theta_2$ is $pE(\cos\theta_1 - \cos\theta_2)$. For a dipole of moment 4 × 10⁻⁹ C m in a field of 5 × 10⁴ N C⁻¹, turning it from alignment through 180° costs $2pE = 4 \times 10^{-4}\ \text{J}$.

Conductors in electrostatics

Free electrons in a conductor move until they cancel any field inside. In equilibrium, five things hold.

Electrostatic shielding. The field inside a cavity of a conductor, with no charge in the cavity, is zero whatever fields or charges are on the outside. Sensitive instruments are put inside metal boxes for this reason.

Beyond the textbook. For the exam: a metal enclosure screens the inside from external electric fields. In reality: this is why you are safe inside a closed car during a lightning storm. The charge flows around the metal body and the field inside is nearly zero (the effect is called a Faraday cage). The same law explains lightning conductors: on a pointed conductor the charge density is greatest at the sharpest point, so a pointed lightning rod leaks charge into the air readily and gives lightning a safe path to the ground. Why it matters: the same reasoning gives you the charge distribution on two connected spheres, which is the example below.

Worked example: two connected spheres. Two conducting spheres of radii 4 cm and 8 cm are joined by a thin wire. They carry a total charge of 60 nC. Find the charge on each, their common potential, and the ratio of the surface charge densities.

The common wrong step: sharing the charge equally. It is the potential, not the charge, that equalises. The smaller sphere ends up with less charge but a higher surface density (and a stronger field at its surface), since $\sigma \propto 1/R$.

Dielectrics

A dielectric is an insulator with no free charges. In an external field its molecules still respond. Non-polar molecules (such as methane) acquire an induced dipole moment. Polar molecules (such as water) already have permanent dipoles that line up with the field. Either way the dielectric becomes polarised, and the polarisation P is the dipole moment per unit volume. For a linear dielectric,

$P = \varepsilon_0 \chi_e E$, where $\chi_e$ is the dimensionless electric susceptibility

The bound charges that appear on its surfaces oppose the applied field, so the net field inside is smaller by the dielectric constant $K = 1 + \chi_e$. That is, $E = E_0/K$. A conductor behaves like a dielectric with K very large, since it cancels the field completely.

Capacitors

Two conductors separated by an insulator form a capacitor. Charge +Q on one and −Q on the other produces a potential difference V, proportional to Q. Their ratio is the capacitance

$C = Q/V$, unit farad ($1\ \text{F} = 1\ \text{C V}^{-1}$); practical sizes are µF, nF and pF

C depends only on the geometry of the conductors and on the medium between them. It does not depend on Q or V. A single isolated spherical conductor of radius R has $C = 4\pi\varepsilon_0 R = R/(9 \times 10^{9})$ in SI units.

Beyond the textbook. For the exam: capacitors come in µF and pF. In reality: one farad is enormous. An isolated sphere would need a radius of 9 × 10⁹ m, about thirteen times the radius of the Sun, to have $C = 1$ F. Even the whole Earth has only about 700 µF. Supercapacitors reach farads by packing a huge internal surface area with the plates only nanometres apart. Why it matters: it tells you why the practical capacitors in numericals are measured in microfarads and picofarads.

Parallel plate capacitor. Two plates of area A, separated by d (small compared with the plate size), carry ±Q. The field between them is $E = \sigma/\varepsilon_0 = Q/(\varepsilon_0 A)$, and the potential difference is $V = Ed$. Therefore

$$C = \dfrac{\varepsilon_0 A}{d}$$

Bringing the plates closer or making them larger increases C. Air breaks down at a field of about 3 × 10⁶ V m⁻¹ (the dielectric strength), so a 1 mm air gap sparks at roughly 3000 V.

With a dielectric. Fill the gap completely with a dielectric of constant K. For the same charge the field falls to $E_0/K$, so the potential difference falls by K and the capacitance rises: $C = K\varepsilon_0 A/d = KC_0$. In general $K = C/C_0$. If a slab of thickness t is placed in a gap d, the effective gap is

$$d - t + \dfrac{t}{K}$$

since the slab behaves like a thickness of only t/K of air. A 3 mm slab with $K = 3$ in a 5 mm gap makes the effective gap $5 - 3 + 1 = 3$ mm, so the capacitance is $5/3$ times its original value.

Worked example: inserting a dielectric, battery on and off. A parallel plate capacitor has plate area 100 cm² and gap 1 mm, in air. It is charged by a 12 V battery. A dielectric slab of K = 5 then completely fills the gap, (a) with the battery still connected, (b) after the battery is removed. Find Q, V and the stored energy in each case. ($\varepsilon_0 = 8.85 \times 10^{-12}\ \text{C}^2\ \text{N}^{-1}\ \text{m}^{-2}$.)

The common wrong step: using the same rule in both cases. With the battery on, the voltage is held fixed and charge flows in from the battery (Q grows by K). With the battery off, the charge is trapped and the voltage falls by K. In case (a) the energy rises by a factor K, because the battery does work. In case (b) it falls by a factor K, because the field does work in pulling the slab in.

Combinations. In series the same charge Q is on every capacitor and the voltages add: $1/C = 1/C_1 + 1/C_2 + \cdots$, so the total is smaller than the smallest. In parallel the same voltage is across each and the charges add: $C = C_1 + C_2 + \cdots$. Note that this is the opposite of the resistor rule.

Worked example: capacitors in series. Capacitors of 4 µF and 12 µF are connected in series across 16 V. Find the equivalent capacitance, the charge, the voltage across each, and the energy stored.

The common wrong step: assuming both capacitors get 16 V, or each 8 V. In series the smaller capacitor takes the larger share of the voltage, because $V = Q/C$ for the same $Q$.

Energy stored in a capacitor

Charging a capacitor means moving charge against the growing voltage between the plates. The first bit costs almost nothing, and the last bit costs the full V, so the average cost is V/2 per unit charge. That is why

$$U = \dfrac{1}{2}QV = \dfrac{1}{2}CV^2 = \dfrac{Q^2}{2C}$$

Where is the energy? In the field between the plates. Per unit volume it is the energy density

$$u = \dfrac{1}{2}\varepsilon_0 E^2$$

and this holds for any electric field in vacuum, not only for a parallel plate capacitor.

Pause and predict: a 6 µF capacitor charged to 100 V is connected to an uncharged 3 µF capacitor. What happens to the total energy?

Check your answer: charge is conserved, 600 µC, but now shared across 9 µF, so the common voltage is $600/9 = 66.7$ V. The initial energy is $1/2 \times 6 \times 10^{-6} \times 100^2 = 30$ mJ and the final energy is $1/2 \times 9 \times 10^{-6} \times 66.7^2 = 20$ mJ. A third of the energy (10 mJ) disappears, as heat and radiation in the connecting wire. Charge is conserved but energy is not.

Exam traps

Formula sheet

60-second revision

Where this chapter connects

• Before this: Electric Charges and Fields gives the field and Gauss's law that this chapter turns into potential, and Work, Energy and Power gives the idea of a conservative force.
• The same idea elsewhere: Gravitation has the same potential energy pattern, with mass in place of charge.
• Class XII link: Current Electricity builds on potential difference as the driver of current, Alternating Current uses capacitors in circuits, and Dual Nature of Radiation and Matter uses the electron volt and stopping potentials.