Chapter 2 Electric Fields
Recommended reading: Tipler & Mosca Chapters 21,22,23 and some of 24 (please note that we are not covering dielectrics).
2.1 Coulomb’s Law
Recommended reading: Tipler & Mosca 21-3
Coulomb’s Law gives the magnitude and direction of the forces between charges.
Figure 2.1: Two charges, q_1 and q_2, separated by a distance of r_12, with the unit vector r_12 pointing in the direction from q_1 to q_2.
The force on \(q_2\) due to \(q_1\) is given by:
\[\begin{equation} \tag{2.1} \mathbf{F}_{12}= \frac{1}{4\pi \epsilon_0} \frac{q_1 q_2}{r_{12}^2} \hat{\mathbf{r}}_{12} \end{equation}\]
The force on \(q_2\) is directed away from \(q_1\) if \(q_1 q_2\) is positive (i.e. when the charges have the same sign - like charges repel). If \(q_1 q_2\) is negative (charges have different signs), the force on \(q_2\) is directed towards \(q_1\) (unlike charges attract).
2.2 The electric field
The electric field (\(\mathbf{E}\)) is defined in terms of the force on a test charge \(q_0\). The force on \(q_0\) due to another charge \(q\) is
\[\begin{equation} \tag{2.2} \mathbf{F} = \frac{1}{4\pi \epsilon_0} \frac{q q_0}{r_{12}^2} \hat{\mathbf{r}}_{12} \end{equation}\]
The electric field is the force divided by the magnitude of the test charge – the field is independent of the charge used to test it.
\[\begin{equation} \tag{2.3} \mathbf{E} = \frac{\mathbf{F}}{q_0} = \frac{1}{4\pi \epsilon_0} \frac{q}{r_{12}^2} \hat{\mathbf{r}}_{12} \end{equation}\]
2.3 Principle of superposition
What if there are several charges?
Have a look at the diagram below:
Figure 2.2: Three charges, 1, 2 and 3, where charges 1 and 3 are negative and charge 2 is positive. The forces shown are the forces exerted on charge 1 by charge 2 (F_21) and charge 3 (F_31), and the total resultant force on charge 1, F_R.
The principle of superposition says that the resultant force \(\mathbf{F}_R\) on charge 1 due to charge 2 and charge 3 is simply the vector sum of the forces due to the individual charges.
The principle of superposition can be used to calculate the electric field \(\mathbf{E}\) for a group of charges by summing the fields due to all charges present.
\[\begin{equation} \tag{2.4} \mathbf{E} = \mathbf{E}_1 + \mathbf{E}_2 + \mathbf{E}_3 + ... = \sum_i \mathbf{E}_i \end{equation}\]
2.4 Electric dipole field
Recommended reading: Tipler & Mosca 21-4
An electric dipole is a combination of a positive and negative charge, equal in magnitude, a small distance from each other. The field for an electric dipole can be calculated by summing the field due to the two charges.
Figure 2.3: The electric field associated with a dipole.
Consider the diagram of the dipole below:
Figure 2.4: A dipole represented in the x-y plane, where d is the distance between the charges.
The magnitude of the field at point \(P\) is given by
\[\begin{equation} \tag{2.5} E = \frac{1}{4 \pi \epsilon_0} \frac{qd}{\left( x^2 + \left( \frac{d}{2} \right)^2 \right)^{\frac{3}{2}}} \end{equation}\]
and it is in the \(-\hat{\mathbf{j}}\) (or \(-\hat{\mathbf{y}}\)) direction.
2.4.1 Electric dipole moment
Definition: The electric dipole moment, \(\mathbf{p}\), is the product of the charge and the vector displacement from the negative charge to the positive charge in a dipole.
Figure 2.5: A diagram of an electric dipole, showing the vector quantities f{d} and p, which are the length of the dipole and the dipole moment respectively.
\[\begin{equation} \tag{2.6} \mathbf{p} = q \mathbf{d} \end{equation}\]
The electric field of the dipole (Equation (2.5)) can therefore be expressed (as a vector) in terms of the dipole moment, as follows:
\[\begin{equation} \tag{2.7} \mathbf{E} = \frac{1}{4 \pi \epsilon_0} \frac{\mathbf{p}}{\left( x^2 + \left( \frac{d}{2} \right)^2 \right)^{\frac{3}{2}}} \end{equation}\]
In the limit of \(x \gg d\) (in other words, when we are at a distance \(x\) from the dipole that is much larger than the size of the dipole \(d\)) the electric field due to the dipole can be reduced to:
\[\begin{equation} \tag{2.8} \mathbf{E} = \frac{1}{4\pi \epsilon_0} \frac{\mathbf{p}}{x^3} \end{equation}\]
2.4.2 Dipoles in external electric fields
Consider a dipole in a uniform electric field. The force on each charge has magnitude \(qE\), but since these forces are in opposite directions there is no net force on the dipole. There is, however a torque about the centre of the dipole.
Figure 2.6: Representation of a dipole placed in an external uniform electric field. θ is the angle the dipole takes to the direction of the external electric field and F is the force on each charge of the dipole due to the external field.
The torque about the centre of the dipole is
\[\begin{equation} \tag{2.9} \tau = 2 F \frac{d}{2} \sin\theta = q E d \sin\theta = |p|E \sin\theta \end{equation}\]
The direction of the torque is perpendicular to the page, so it can be represented in vector form as \(\boldsymbol{\tau} = \mathbf{p} \times \mathbf{E}\).
The torque will cause the dipole to rotate and align itself with the electric field as shown here:
We can use the work done by the field to determine what is the minimum energy configuration (although it should be fairly obvious). The work done is the integral of the product of the torque and the angle turned through:
\[\begin{equation} \tag{2.10} \begin{array}{rcll} W &=& \int_{\theta_0}^{\theta} |\tau| \mathrm{d}\theta \\ &=& \int_{\theta_0}^{\theta} pE \sin\theta \mathrm{d}\theta \\ &=& [-pE \cos\theta]_{\theta_0}^{\theta} \end{array} \end{equation}\]
The change in potential energy is \(\Delta U = W\), hence: \[\begin{equation} \tag{2.11} \Delta U = U(\theta_0) - U(\theta) = pE(\cos\theta_0 - cos\theta) \end{equation}\]
The zero of potential energy \(U(\theta_0)\) can be chosen to be anywhere, so we can choose it to correspond to \(\theta_0 = 90^{\circ}\) in which case \(U = -pE \cos\theta\). This energy can be expressed in vector form as \(U = - \mathbf{p} \cdot \mathbf{E}\).
Not surprisingly, the energy is a minimum when the dipole is aligned with the field at which point the torque will be zero.
2.5 Continuous charge distributions
Recommended reading: Tipler & Mosca Chapter 22
Charges are discrete i.e. all charges sit on point like particles but if we have a large number of changes they can be treated as a continuous charge distribution. Continuous charge distributions can be described by linear, surface or volume charge densities. To use Coulomb’s Law to calculate the electric field in these cases you may need to integrate using the charge density.
2.6 Electric flux & Gauss’s Law
Flux has been found to be a concept that’s often misunderstood when discussing electric fields.
But in the case of the electric field from static charges nothing is “flowing”. You can think of it as the number of field lines passing through a unit of area. Alternatively consider an analogy of a light source – a source of photons.
Since photons travel in straight lines, the number of photons passing through area \(A_1\) in unit time is the same as that passing through area \(A_2\), i.e. the flux of photons is the same, \(I_1 A_1 = I_2 A_2\) – where \(I_{1,2}\) is photon intensity. The intensity of photons is proportional to \(\frac{1}{A} \propto \frac{1}{r^2}\) so the photon field depends on \(\frac{1}{r^2}\). Replace the light source with a source of electric field e.g. a point charge. The field starts at the point charge and spreads out. The amount of field doesn’t increase as we move away from the charge because there is no source of electric field.
Consider the field in and out of the volume enclosed between the two areas (a truncated cone). The total field in and out is the surface integral of the component of the field normal to the surfaces. On the curved faces, the normal component of \(\mathbf{E}\) is zero. On the spherical faces, \(A_1\) and \(A_2\), the field is normal. If flux into the volume is negative, and flux out is positive \(\mathbf{E}\) decreases with \(\frac{1}{r^2}\) and the area increases with \(r^2\). So the fluxes through the two faces, \(A_1\) and \(A_2\), are equal and opposite.
If a surface is tilted at an angle \(\theta\) the field normal to the surface is \(\mathbf{E}_n = \mathbf{E} \cos\theta\). To determine the flux through the surface we need ∯ \(E \cos\theta \mathrm{d} A\), where the integral is over the surface. We can see that this flux is still the same as that through \(A_1\) and \(A_2\) because all the flux that impinges on \(A_2\) also impinges on \(A_3\).
Consider a spherical surface with a point charge \(q\) at the centre:
The net flux is non-zero as \(\mathbf{E}\) is pointing outwards over the whole spherical surface. Because of the symmetry the magnitude of \(\mathbf{E}\) is consistent across the surface and \(\mathbf{E}\) is always perpendicular to the surface.
Total flux of \(\mathbf{E}\) is \(\Phi_{\mathbf{E}} = E_n \times Area\). Therefore \[\begin{equation} \tag{2.12} \Phi_E = \frac{q}{4\pi\epsilon_0 r^2} \times 4\pi r^2 = \frac{q}{\epsilon_0} \end{equation}\]
This result is independent of the radius of the sphere – in fact, it is independent of the shape of the surface enclosing the charge.
2.7 Gauss’s Law
(Note: In Equations (2.13), (2.14) and (2.15), \(\iint_S\) represents the closed surface integral, which is denoted elsewhere in the text as ∯.)
We can calculate the electric field from any charge distribution with Coulomb’s Law. Gauss’ Law in the integral form which we’ll use here can calculate fields in some highly symmetric cases. Gauss’ Law states that for any closed surface \(S\):
\[\begin{equation} \tag{2.13} \iint_S{E_n} \mathrm{d} \mathbf{S} = 0 \end{equation}\]
if no charge is enclosed and
\[\begin{equation} \tag{2.14} \iint_S{E_n} \mathrm{d} \mathbf{S} = \frac{q}{\epsilon_0} \end{equation}\]
if charge \(q\) is enclosed. Or in words: the magnitude of the electric field normal to the surface integrated over the whole of the surface is equal to the charge enclosed divided by \(\epsilon_0\).
More generally, the integral form of Gauss’s Law is \[\begin{equation} \tag{2.15} \iint_S \mathbf{E} \cdot \mathrm{d} \mathbf{S} =\frac{Q}{\epsilon_0} \end{equation}\]
\(\mathrm{d} \mathbf{S}\) is a vector normal to the surface with magnitude equal to the size of an element of area. \(\mathbf{E} \cdot \mathrm{d} \mathbf{S}\), which is a scalar product, extracts the component of the electric field normal to the surface and multiplies it by the size of an element of the area. \(Q\) is the total charge enclosed within \(S\), where \(Q = \sum_i q_i\) where the sum runs over all charges inside \(S\) (where \(S\) is a closed surface enclosing a volume \(V\)).
You can also think in terms of field lines the total number of field lines leaving the surface is proportional to the total number of charges inside the surface. If there are positive and negative charges then some field lines will go in and some out. if the amount of positive and negative charge is equal these two contributions cancel. Note that this doesn’t mean that the is no field anywhere at the surface but that the contributions of positive and negative flux over the whole surface cancel. The integral form of Gauss’s Law can be used to find the electric field for symmetrical systems of charges. We’ll look at three classic examples:
Field due to a line of charge, linear charge density \(\lambda\) C/m
Field due to an infinite plane of charge, surface charge density \(\sigma\) C/m\(^2\)
Field due to a solid sphere of uniformly distributed charge \(Q\)
2.7.1 Field inside a hollow object
Consider a spherical shell with charge \(Q\). By symmetry, the field is radial in all directions.
Applying Gauss’s Law with a spherical Gaussian surface outside the shell, as shown in , gives: \[\begin{equation} \tag{2.16} E_r = \frac{Q}{4\pi\epsilon_0 r^2} \end{equation}\]
i.e. the same as point charge or a solid sphere of charge.
Figure 2.7: A spherical shell (dark red area) of charge Q and radius r producing an electric field E (light red arrows). The dotted line outside the shell can be selected as a Gaussian surface.
Applying Gauss’s Law with a spherical Gaussian surface inside the shell gives \(E_r = 0\) because there is no charge inside the surface.
There is no electric field inside a uniform shell of charge.
In a conductor, charges are free to move and as like charges repel excess charge will move to the outside surface of an isolated conductor.
Figure 2.8: The cross-section of a conductor, where the red circles represent the excess charge which will move to the outside surface of the conductor. The dotted line can be selected as a Gaussian surface to show that the field inside the conductor is zero (because there is no (net) charge enclosed in this Gaussian surface).
Consider a Gaussian surface just inside a conductor as shown in the diagram above. The electric field is zero everywhere inside the conductor – because all the charges are on the outside. The flux through the Gaussian surface must therefore be zero.
This applies if there is a cavity inside the conductor and so inside any cavity in a conductor, the electric field is zero. This principle is used to build Faraday Cages to shield sensitive electronics.
The field at the surface of a conductor is always perpendicular to the surface. If this wasn’t the case, there would be a component of the field tangential to the surface. The charges in the conductor would move across the surface under the influence of the tangential component of the field until that component was zero. Therefore the field is always perpendicular to the surface.
2.8 Electric potential
The electric potential due to a positive charge is positive. The electric potential due to a negative charge is negative. The potential difference between two points, \(a\) and \(b\), is
\[\begin{equation} \tag{2.17} \Delta V = V_b - V_a =\frac{(U_b - U_a)}{q_0} \end{equation}\]
The electric (or electrostatic) potential, often just called potential, is defined as the potential energy per unit (test) charge.
Consider bringing an infinitesimal test charge (\(q_0\)) from infinity into a region containing a system of charges. If \(U\) is the final potential energy of the charge the electric potential is given by \(V = U/q_0\).
The change in potential energy in going from point \(a\) to point \(b\) is given by the integral along the path from \(a\) to \(b\) of the work done on the charge:
\[\begin{equation} \tag{2.18} \Delta U = - W_{ab} = -\int_a^b \mathbf{F} \cdot \mathrm{d}\mathrm{l} = -q_0 \int_a^b \mathbf{E} \cdot \mathrm{d}\mathrm{l} \end{equation}\]
We know that \[\begin{equation} \tag{2.19} \frac{\Delta U}{q_0} = V_b - V_a \end{equation}\]
hence
\[\begin{equation} \tag{2.20} \Delta V_{ab} = -\int_a^b \mathbf{E} \cdot \mathrm{d}\mathrm{l} \end{equation}\]
So the change in electrical potential is the line integral of the electric field along the path from \(a\) to \(b\). We can choose the reference point of the potential to be where we like but normally select infinity. So
\[\begin{equation} \tag{2.21} V_b = -\int_\infty^b \mathbf{E} \cdot \mathrm{d}\mathrm{l} \end{equation}\]
The potential at a distance \(r\) from a point charge is
\[\begin{equation} \tag{2.22} \begin{array}{rcll} V(r) &=& -\int_\infty^r \frac{1}{4\pi\epsilon_0} \frac{q}{r^2} \mathrm{d} r \\ &=& \frac{q}{4\pi\epsilon_0} \left[ \frac{1}{r} - \frac{1}{\infty} \right] \\ &=& \frac{q}{4\pi\epsilon_0} \frac{1}{r} \end{array} \end{equation}\]
Note: The dependence of the potential for a point charge on distance is \(\frac{1}{r}\) and the force depends on \(\frac{1}{r^2}\) because the electric field is the gradient of the potential.