Electric Fields
Rub a balloon on your jumper and it can lift your hair without touching it: a charged object fills the space around it with an electric field. In this topic we define field strength, use Coulomb's law for the force between two charges, meet electric potential, and see how field and potential describe the same thing in two different ways. These ideas run straight on into capacitors, particle accelerators and the deflection of electron beams.
Electric fields and field strength
Electric field
An electric field is a region in which a charged particle experiences a force. It is a force field, like a gravitational field, except that it acts on charge instead of mass.
To measure how strong a field is at some point, imagine placing a small positive "test" charge there and measuring the force on it. Twice the charge feels twice the force, so the useful quantity is the force on each coulomb: the electric field strength.
Electric field strength
Electric field strength
The electric field strength at a point is the force per unit charge on a small positive test charge placed at that point. It is a vector, pointing in the direction of the force on a positive charge, and is measured in N C⁻¹.
- A positive charge is pushed along the field direction.
- A negative charge, such as an electron, is pushed the opposite way to the field.
- The test charge has to be small so that it doesn't push the charges that make the field around and change the very thing you're measuring.
An electron is in a uniform electric field of strength \(5.0 \times 10^{4}\ \text{N C}^{-1}\). Calculate the force on it and its acceleration, and compare the force with its weight.
Show solution
\[ F = EQ = 5.0\times10^{4} \times 1.60\times10^{-19} = 8.0 \times 10^{-15}\ \text{N} \] \[ a = \frac{F}{m} = \frac{8.0\times10^{-15}}{9.11\times10^{-31}} = 8.8 \times 10^{15}\ \text{m s}^{-2} \]The weight is \(mg = 9.11\times10^{-31} \times 9.81 = 8.9 \times 10^{-30}\ \text{N}\), about \(10^{15}\) times smaller than the electric force. That's why gravity is ignored for electrons and ions in electric fields. The force acts in the opposite direction to the field, because the electron is negative.
Coulomb's law
Two point charges push each other apart if they are alike and pull together if they are opposite. Coulomb found that the size of the force depends on the charges and the distance between them in a very simple way.
Coulomb's law
The force between two point charges is proportional to the product of the charges and inversely proportional to the square of the distance between them.
Coulomb's law
The constant at the front is often written \(k = \dfrac{1}{4\pi\varepsilon_0} = 8.99 \times 10^{9}\ \text{N m}^2\,\text{C}^{-2}\). It is enormous, which tells you that a coulomb is a huge amount of charge: two charges of 1 C held 1 m apart would push on each other with about \(9 \times 10^9\ \text{N}\). Everyday static charges are microcoulombs or nanocoulombs.
- Sign: if \(Q_1Q_2\) is positive (like charges), \(F\) is positive and the force is repulsive. If it is negative (unlike charges), the force is attractive.
- Newton's third law: each charge feels the same size of force, in opposite directions, even if one charge is much bigger than the other.
- Inverse square: double the distance and the force becomes \(\tfrac14\); triple it and the force becomes \(\tfrac19\).
Three classic slips: forgetting to square \(r\); using the distance between the surfaces of two spheres instead of their centres; and not converting μC (\(10^{-6}\) C) or nC (\(10^{-9}\) C). Do the sign separately: work out the size of the force, then decide "attract" or "repel" from the charges.
In a simple model of a hydrogen atom, the electron orbits the proton at a distance of \(5.3 \times 10^{-11}\ \text{m}\). (a) Calculate the electric force between them. (b) Compare it with the gravitational force, \(3.6\times10^{-47}\ \text{N}\). (c) The electric force provides the centripetal force. Find the electron's speed.
Show solution
(a) Both charges have size \(e = 1.60\times10^{-19}\ \text{C}\):
\[ F = \frac{8.99\times10^{9} \times (1.60\times10^{-19})^2}{(5.3\times10^{-11})^2} = \frac{2.30\times10^{-28}}{2.81\times10^{-21}} = 8.2 \times 10^{-8}\ \text{N} \]The charges are opposite, so the force is attractive.
(b) \(8.2\times10^{-8} / 3.6\times10^{-47} \approx 2 \times 10^{39}\). Gravity is completely negligible inside atoms.
(c) Using circular motion, \(F = mv^2/r\):
\[ v = \sqrt{\frac{Fr}{m}} = \sqrt{\frac{8.2\times10^{-8} \times 5.3\times10^{-11}}{9.11\times10^{-31}}} = 2.2 \times 10^{6}\ \text{m s}^{-1} \]That is under 1% of the speed of light, so the non-relativistic formulas are fine here.
The field of a point charge
Combine Coulomb's law with the definition of field strength and you get the field at any distance from a single charge.
Place a small positive test charge \(q\) a distance \(r\) from a point charge \(Q\). Coulomb's law gives the force on it:
\[ F = \frac{Qq}{4\pi\varepsilon_0 r^2} \]Field strength is force per unit charge, so divide by \(q\):
\[ E = \frac{F}{q} = \frac{Qq}{4\pi\varepsilon_0 r^2 \, q} \quad\Rightarrow\quad \boxed{E = \frac{Q}{4\pi\varepsilon_0 r^2}} \]The test charge cancels: the field belongs to \(Q\) alone, whether or not anything is there to feel it.
Field strength of a point charge
This is a radial field: the field lines point straight out from a positive charge (and straight in towards a negative one), and they spread apart as \(r\) increases, showing that the field weakens with the square of the distance. Outside a charged sphere, the field is the same as if all the charge were concentrated at its centre, so you measure \(r\) from the centre.
The dome of a Van de Graaff generator is a sphere of radius 0.15 m carrying a charge of \(+2.0\ \mu\text{C}\). Calculate the field strength at its surface and 0.30 m from its centre.
Show solution
\[ E_{0.15} = \frac{8.99\times10^{9} \times 2.0\times10^{-6}}{0.15^2} = 8.0 \times 10^{5}\ \text{N C}^{-1} \] \[ E_{0.30} = \frac{8.99\times10^{9} \times 2.0\times10^{-6}}{0.30^2} = 2.0 \times 10^{5}\ \text{N C}^{-1} \]Doubling the distance from the centre quarters the field. Notice that 0.30 m from the centre is only 0.15 m from the surface: always measure from the centre.
Electric potential
Field strength tells you the force on each coulomb. Potential tells you the energy of each coulomb. Imagine pushing a small positive charge in from very far away towards a positive charge \(Q\). It is repelled, so you must do work on it, and that work is stored as electric potential energy. The work done per coulomb is the potential at the point where you stop.
Electric potential
The electric potential at a point is the work done per unit charge in bringing a small positive test charge from infinity to that point. It is a scalar, measured in volts (1 V = 1 J C⁻¹), and is zero at infinity.
Potential in a radial field
- Near a positive charge the potential is positive (you had to push your test charge in) and falls towards zero as \(r\) increases.
- Near a negative charge the potential is negative (the test charge is pulled in, so energy is given out) and rises towards zero as \(r\) increases.
- Notice the powers: \(V \propto 1/r\) but \(E \propto 1/r^2\). Double the distance and the potential halves but the field quarters.
- The potential difference between two points is the work done per unit charge moving between them, so the work done moving charge \(Q\) through a p.d. \(\Delta V\) is \(W = Q\,\Delta V\), exactly as in Unit 2.
An equipotential is a line (or surface) joining points at the same potential. Moving a charge along an equipotential takes no work, so the field can have no component along it: equipotentials are always at right angles to field lines. Around a point charge they are concentric spheres. Drawn at equal steps of potential, they get further apart as you move out, because the field is getting weaker.
How field and potential are related
Field and potential are two descriptions of the same field: one tells you about forces, the other about energies. The link between them is the gradient of the potential.
Move a small positive charge \(q\) a short distance \(\Delta r\) in the direction of the field \(E\). The field pushes it with force \(qE\), so the field does work on it:
\[ W = qE\,\Delta r \]That energy comes out of the charge's electric potential energy, which falls by \(qE\,\Delta r\). Per unit charge, the potential falls by \(E\,\Delta r\):
\[ \Delta V = -E\,\Delta r \]Rearrange:
\[ \boxed{E = -\frac{\Delta V}{\Delta r}} \]The field strength is the gradient of the \(V\)–\(r\) graph with its sign reversed. The minus sign means the field points "downhill", from high potential to low potential.
Reading the two graphs
- The gradient of a \(V\)–\(r\) graph at any point is \(-E\). Where the curve is steepest, the field is strongest.
- Equipotentials close together mean a steep potential gradient, which means a strong field.
- Field lines always point from higher potential to lower potential.
- Units check: the gradient is in V m⁻¹, and \(1\ \text{V m}^{-1} = 1\ \text{J C}^{-1}\,\text{m}^{-1} = 1\ \text{N m C}^{-1}\,\text{m}^{-1} = 1\ \text{N C}^{-1}\).
(a) Calculate the potential at the surface of the Van de Graaff dome from the previous example (radius 0.15 m, charge \(+2.0\ \mu\text{C}\)) and 0.30 m from its centre. (b) A proton moves from a point where the potential is 60 kV to a point where it is 20 kV. Starting from rest, how fast is it moving?
Show solution
(a)
\[ V_{0.15} = \frac{8.99\times10^{9} \times 2.0\times10^{-6}}{0.15} = 1.2 \times 10^{5}\ \text{V}, \qquad V_{0.30} = 6.0 \times 10^{4}\ \text{V} \]Doubling the distance halved the potential but quartered the field. For a point charge, \(E = V/r\): \(1.2\times10^5/0.15 = 8.0\times10^5\ \text{N C}^{-1}\), matching the earlier answer.
(b) The proton falls through a potential difference of \(60 - 20 = 40\ \text{kV}\). The work done on it by the field becomes kinetic energy:
\[ E_k = Q\,\Delta V = 1.60\times10^{-19} \times 4.0\times10^{4} = 6.4 \times 10^{-15}\ \text{J} \] \[ v = \sqrt{\frac{2E_k}{m}} = \sqrt{\frac{2 \times 6.4\times10^{-15}}{1.67\times10^{-27}}} = 2.8 \times 10^{6}\ \text{m s}^{-1} \]Uniform fields between parallel plates
Connect two parallel metal plates to a supply and one becomes positive and the other negative. Between them (away from the edges) the field lines are parallel, equally spaced and at right angles to the plates: the field is uniform, the same strength and direction everywhere. Near the edges the lines bulge outwards; these are called edge effects.
Move a positive charge \(Q\) from the negative plate to the positive plate, a distance \(d\), against the field. The force needed is \(F = EQ\), which is constant because the field is uniform. The work done is
\[ W = Fd = EQd \]By the definition of potential difference, moving charge \(Q\) through p.d. \(V\) takes work
\[ W = QV \]The two must be equal. \(Q\) cancels:
\[ EQd = QV \quad\Rightarrow\quad \boxed{E = \frac{V}{d}} \]So the unit of field strength can also be written V m⁻¹. This is just \(E = -\Delta V/\Delta r\) for a field whose potential falls steadily from one plate to the other.
Uniform field between parallel plates
The potential falls at a steady rate from the positive plate to the negative plate, so the equipotentials are planes parallel to the plates and equally spaced. Halfway across, the potential is halfway between the two plate potentials. Notice that \(E\) depends only on \(V\) and \(d\), not on the area of the plates or where between them you are.
Charged particles crossing a uniform field
Fire an electron horizontally between horizontal plates. The force \(eE\) is constant and vertical, so this is exactly like a projectile under gravity: the horizontal velocity stays constant and there is a constant vertical acceleration \(a = eE/m\). The path is a parabola. Once the electron leaves the plates, the force stops and it carries on in a straight line.
With horizontal speed \(v_0\) and plate length \(L\), the time between the plates is \(t = L/v_0\), and the sideways deflection at the exit is \(y = \tfrac12 a t^2 = \dfrac{eV L^2}{2 m d\, v_0^2}\). Bigger \(V\) bends it more; a faster electron spends less time in the field and bends less.
An electron travelling at \(3.0\times10^{7}\ \text{m s}^{-1}\) enters the space midway between two parallel plates 8.0 cm long and 2.0 cm apart, moving parallel to them. The p.d. between the plates is 150 V. Calculate (a) the field strength, (b) the acceleration of the electron, (c) its deflection as it leaves the plates and (d) the angle its path then makes with its original direction.
Show solution
(a) \[ E = \frac{V}{d} = \frac{150}{0.020} = 7500\ \text{V m}^{-1} \]
(b) \[ a = \frac{eE}{m} = \frac{1.60\times10^{-19} \times 7500}{9.11\times10^{-31}} = 1.32 \times 10^{15}\ \text{m s}^{-2} \]
(c) Horizontally the speed is constant: \(t = L/v_0 = 0.080/3.0\times10^{7} = 2.67 \times 10^{-9}\ \text{s}\). Vertically it starts from rest:
\[ y = \tfrac12 a t^2 = \tfrac12 \times 1.32\times10^{15} \times (2.67\times10^{-9})^2 = 4.7 \times 10^{-3}\ \text{m} = 4.7\ \text{mm} \]That is less than the 10 mm gap to the plate, so the electron gets through.
(d) Vertical velocity at the exit: \(v_y = at = 1.32\times10^{15} \times 2.67\times10^{-9} = 3.5\times10^{6}\ \text{m s}^{-1}\).
\[ \theta = \tan^{-1}\!\left(\frac{3.5\times10^{6}}{3.0\times10^{7}}\right) = 6.7^\circ \]Drawing field lines and equipotentials
Exam questions often ask you to sketch or interpret field patterns. The rules are few:
- Field lines start on positive charges and end on negative charges (or go off to infinity). The arrow shows the direction of the force on a positive charge.
- Field lines never cross: the field can only point one way at any point.
- The closer the lines, the stronger the field. Parallel, equally spaced lines mean a uniform field.
- Field lines meet the surface of a conductor at 90°.
- Equipotentials are always perpendicular to field lines. Drawn at equal steps of potential, they are closer together where the field is stronger.
A field line is not the path a charge will follow. It shows the direction of the force, and so of the acceleration. A charge that is already moving, or one released in a curved field, will not stay on a field line, for the same reason a thrown ball doesn't move in the direction of gravity.
Electric and gravitational fields side by side
| Electric | Gravitational | |
|---|---|---|
| Acts on | charge | mass |
| Field strength | \(E = F/Q\), N C⁻¹ | \(g = F/m\), N kg⁻¹ |
| Force between points | \(Q_1Q_2/4\pi\varepsilon_0r^2\): attract or repel | \(Gm_1m_2/r^2\): always attract |
| Potential | \(Q/4\pi\varepsilon_0r\): positive or negative | \(-Gm/r\): always negative |
| Uniform field | between parallel plates | close to the Earth's surface |
Both are inverse-square fields, which is why the maths is identical. You'll meet the gravitational versions on the Gravitational Fields page in Unit 5.
Quick checks
The distance between two point charges is doubled. What happens to the force between them?
\(F \propto 1/r^2\), so doubling \(r\) gives \(1/2^2 = 1/4\) of the force. "Halves" would be the answer for the potential, which goes as \(1/r\).
Which of these is a unit of electric field strength?
From \(E = V/d\), field strength is in V m⁻¹, which is the same as N C⁻¹. J C⁻¹ is the volt, the unit of potential.
The electric field at a point points due east. Which way is the force on an electron placed there?
The field direction is the direction of the force on a positive charge. The electron is negative, so it is pushed the opposite way. Electric forces act on charges whether or not they are moving.
Equipotentials around a point charge are drawn at equal steps of potential. Moving away from the charge, they
The field gets weaker further out, so the potential changes more slowly with distance: you have to go further to drop by the same number of volts.
The p.d. across a pair of parallel plates is kept the same while their separation is doubled. What happens to the force on a charge between them?
\(E = V/d\), so doubling \(d\) halves \(E\), and \(F = EQ\) halves too. It is not an inverse-square effect: the field between plates is uniform.
Point P is midway between a charge \(+Q\) and a charge \(-Q\). Which statement is correct at P?
Potential is a scalar: \(+kQ/r\) and \(-kQ/r\) add to zero. Field is a vector: both charges' fields at P point the same way (away from \(+Q\), towards \(-Q\)), so they add rather than cancel. Potential has no direction at all.