A trekking map of the hills near Manali shows height by contour lines, each joining points at the same height. Where the lines crowd together, the slope is steep and a dropped ball runs fast. Where they spread apart, the ground is nearly flat. Water always runs straight across the lines, never along them.
From potential back to field
In the chapter on Potential Energy you found a force from its potential energy: \(F = -dU/dx\). Divide both sides by the charge. Potential energy per coulomb is the potential \(V\), and force per coulomb is the field \(E\). So the field is minus the slope of the potential.
The same result follows from the work. Move a unit charge a small step \(dx\) along \(x\). The field does work \(E_x\,dx\), so the potential changes by \(dV = -E_x\,dx\). The quantity \(dV/dx\) is the potential gradient. The field is its negative, so it points the way \(V\) falls, and it is strong where \(V\) changes quickly.
A volt per metre and a newton per coulomb are the same unit, because \(1\,\text{V/m} = 1\,\text{J/(C m)} = 1\,\text{N/C}\). In three dimensions each component is minus a partial derivative, the slope along one axis with the other two held fixed. This is the gradient you met for potential energy.
Find the field for (a) \(V = 5x + 3\), (b) \(V = 2x^2 - 3y\) at the point \((1\,\text{m}, 2\,\text{m})\), and (c) \(V = 20/r\) at \(r = 2\,\text{m}\). All are in SI units. In (c), what charge makes this potential?
Take \(E_x = -\partial V/\partial x\), \(E_y = -\partial V/\partial y\), and \(E_r = -dV/dr\) for (c).
(a) \(dV/dx = 5\). (b) \(\partial V/\partial x = 4x\), \(\partial V/\partial y = -3\). (c) \(dV/dr = -20/r^2\).
(a) \(E_x = -5\,\text{V/m}\): \(5\,\text{V/m}\) along \(-x\), the same everywhere. (b) \(E = (-4x, 3)\), which is \((-4, 3)\) at \((1, 2)\), of size \(5\,\text{V/m}\). (c) \(E = 20/r^2 = 5\,\text{V/m}\), outward.
In (c), \(kq = 20\), so \(q = 20/(9 \times 10^9) = 2.2 \times 10^{-9}\,\text{C}\), a positive charge of \(2.2\,\text{nC}\). Its field points outward, the way \(V\) falls.
Along the \(x\) axis the potential rises steadily from left to right. Which way does the electric field point?
The potential along a line is \(V = 4x^2\) volts, with \(x\) in metres. What is \(E_x\) at \(x = 0.5\,\text{m}\), in V/m? Give the sign.
Reading the two graphs
A graph of \(V\) against \(x\) gives the field at a glance: read its slope and change the sign. Going the other way, \(V_B - V_A = -\int_A^B E_x\,dx\), so a change in \(V\) is minus the area under the \(E_x\) graph. Each graph can be drawn from the other.
In the first chip, \(V\) is \(10\,\text{V}\) up to \(x = 2\,\text{m}\), then falls to \(-10\,\text{V}\) at \(4\,\text{m}\) and stays there. So \(E_x\) is zero, then \(-(-20)/2 = +10\,\text{V/m}\), then zero again. In the middle region the field points along \(+x\), the way \(V\) falls. For the point charge, \(V = 20/x\) and \(E_x = 20/x^2\) fall together. For the uniform field, the straight fall of \(V\) gives a constant \(E_x\).
The field along the \(x\) axis is \(E_x = 5\,\text{V/m}\) from \(x = 0\) to \(2\,\text{m}\), and zero from \(2\,\text{m}\) to \(4\,\text{m}\). The potential at \(x = 0\) is \(20\,\text{V}\). Find the potential at \(2\,\text{m}\) and at \(4\,\text{m}\).
From \(0\) to \(2\,\text{m}\) the area under the graph is \(5 \times 2 = 10\,\text{V}\), so \(V(2) = 20 - 10 = 10\,\text{V}\).
From \(2\,\text{m}\) to \(4\,\text{m}\) the field is zero, so the potential does not change: \(V(4) = 10\,\text{V}\).
The potential falls along the field, since \(dV = -E_x\,dx\) is negative for a step along \(E\).
The uniform field
In a uniform field \(E\) the integral is simply a product. Moving a distance \(d\) along the field lowers the potential by \(Ed\): \(V_A - V_B = Ed\). Moving across the field changes nothing, because the field does no work on a step perpendicular to it. Only the part of a displacement along the field counts.
In Fig 4.1, \(V_A - V_B = 200 \times 0.5 = 100\,\text{V}\), and A is the higher. B and C lie on the same plane, so \(V_B - V_C = 0\). Then \(V_A - V_C = 100\,\text{V}\) as well, although C is farther from A than B is.
Two large parallel plates are \(2\,\text{cm}\) apart. The lower plate is at \(+400\,\text{V}\) and the upper plate at \(0\,\text{V}\). An electron sits midway. Find the field, the force on the electron, and the work done by the field as the electron reaches the lower plate.
\(E = \Delta V/d = 400/0.02 = 2 \times 10^4\,\text{V/m}\). It points from high potential to low, so upward.
\(F = eE = 1.6 \times 10^{-19} \times 2 \times 10^4 = 3.2 \times 10^{-15}\,\text{N}\), downward, because the electron is negative.
The electron moves \(1\,\text{cm}\) along the force: \(W = 3.2 \times 10^{-15} \times 0.01 = 3.2 \times 10^{-17}\,\text{J}\). Check: it crosses \(200\,\text{V}\), and \(e \times 200 = 3.2 \times 10^{-17}\,\text{J}\).
For a point charge the two formulas \(V = kq/r\) and \(E = kq/r^2\) give \(V/E = r\). So measuring both at one point finds the charge. Where \(V = 900\,\text{V}\) and \(E = 9000\,\text{V/m}\), the point is \(10\,\text{cm}\) away, and \(q = Vr/k = 10\,\text{nC}\).
A potential difference of \(50\,\text{V}\) is kept across two parallel plates \(5\,\text{mm}\) apart. What is the field between them, in V/m?
A particle of mass \(8\,\text{g}\) and charge \(2\,\mu\text{C}\) is released from rest in a uniform field of \(2000\,\text{V/m}\). What is its speed after it has moved \(1\,\text{m}\) along the field, in m/s?
Surfaces of equal potential
An equipotential surface joins all the points at one potential. Moving a charge along it changes no potential, so the field does no work. A force does no work only on a displacement perpendicular to it. So the field is perpendicular to the equipotential surface at every point, just as the water on the map runs straight across the contour lines.
The electric field is perpendicular to every equipotential surface and points towards lower potential. No work is done in moving a charge along an equipotential surface, or round any closed path. Two equipotential surfaces of different potential never cross, because a point cannot have two potentials.
Their shapes follow from \(V\). Round a point charge \(V = kq/r\) depends only on \(r\), so the surfaces are spheres centred on the charge. In a uniform field they are planes across the field, as in Fig 4.1. Beside a large charged sheet the field is uniform and perpendicular to the sheet, so they are planes parallel to it.
Where the surfaces are drawn at equal steps of potential, their spacing gives the field. Surfaces \(10\,\text{V}\) apart and \(2\,\text{cm}\) apart mean \(E = 10/0.02 = 500\,\text{V/m}\). Crowded surfaces mean a strong field, as crowded contour lines mean a steep slope.
Play the three chips of Fig 4.2. Along the \(450\,\text{V}\) sphere the work stays zero for all \(31\,\text{cm}\) of path. Straight in to the \(900\,\text{V}\) sphere, \(W = q\,\Delta V = 2 \times 10^{-6} \times 450 = 0.90\,\text{mJ}\). On the round trip the charge travels \(67\,\text{cm}\), yet the work comes back to zero. The work depends on the change in \(V\) alone, never on the length of the path.
Equipotential planes of a uniform field are \(2\,\text{cm}\) apart, at \(30\), \(20\), \(10\) and \(0\,\text{V}\). A charge of \(-2\,\mu\text{C}\) moves from P on the \(30\,\text{V}\) plane to Q on the \(0\,\text{V}\) plane. Find the field and the work done by the field.
\(E = 10\,\text{V}/0.02\,\text{m} = 500\,\text{V/m}\), pointing from the \(30\,\text{V}\) plane towards Q, the way \(V\) falls.
\(W_\text{field} = q(V_P - V_Q) = (-2 \times 10^{-6})(30 - 0) = -6 \times 10^{-5}\,\text{J}\).
The work is negative: the field pushes the negative charge back towards higher potential while it moves the other way.
The next lesson meets the dipole, a pair of charges \(+q\) and \(-q\). Every point of the plane that bisects it at right angles is equally far from both. Their potentials cancel there, so that plane is the equipotential \(V = 0\). Moving a charge anywhere within it takes no work.
Which statements about equipotential surfaces are true?
A charge of \(2\,\mu\text{C}\) is carried slowly from the \(450\,\text{V}\) sphere round a point charge to the \(900\,\text{V}\) sphere. It goes once along a straight line and once along a long spiral. Compare the outside work.
The heart muscle is driven by electric signals, and they set up tiny potential differences all over the body. In 1903 the Dutch physiologist Willem Einthoven recorded them with a very sensitive string galvanometer, between the two arms and a leg. The trace, the electrocardiogram, showed the heart's rhythm. He received the Nobel Prize in 1924. Every ECG in an Indian hospital today still measures potential differences between points on the skin.
- Dropping the minus sign in \(E_x = -dV/dx\).The field points where \(V\) falls, downhill on the V-x graph.
- Reading the field from the value of \(V\) instead of its slope.A large constant \(V\) has no field; a steep \(V\) has a strong one.
- Using the full distance, not the distance along the field, in \(\Delta V = Ed\).Only the part of the displacement along \(E\) changes \(V\).
- Thinking work is needed to move a charge along an equipotential.\(\Delta V = 0\) along it, so \(W = q\,\Delta V = 0\), however long the path.