Static Electric Fields

Abstract

The properties of static electric fields are examined and the concept of electrostatic potential is introduced. It is shown that Gauss’ law leads, in a straightforward way, to Coulomb’s law. The electrostatic properties of a metallic conductor are studied.

Appendices

Questions

• 1. Suggest a physical process by which one may create (a) a static but non-uniform electric field; (b) a uniform but non-static electric field.

• 2. In Gauss’ law in integral form (5.5) the electric field on a closed surface S is associated with the total charge in the interior of S. Will the field on S be affected if we remove all charges in the exterior of this surface?

• 3. (a) Show that the electric field in a region of space where a non-static distribution of charge $\rho \kern0.2em \left(\overrightarrow{r},t\right)$ exists cannot be static. (b) In a region of space the distribution of charge is static. Does the electric field in that region have to be static?

• 4. Justify the principle of superposition for the electric potential: “At any point of space, the electric potential due to a system of charges equals the algebraic sum of potentials due to each charge separately”.

• 5. The electric field is a physical quantity having a uniquely defined value that is measurable at every point of space. Is the same true with regard to the electric potential? Thus, can the potential be regarded as an absolute physical quantity? How about potential difference?

• 6. Starting with Coulomb’s law, derive an expression for the potential of the Coulomb field produced by a point charge q.

• 7. What is an equipotential surface? Show that every such surface intersects normally the electric field lines. Determine the equipotential surfaces of the Coulomb field produced by a point charge q.

• 8. (a) Show that the electrostatic field is conservative and derive an expression for the potential energy of a charge q inside this field. (b) Find the potential energy of a hydrogen atom when the electron is a distance r from the nucleus (proton).

• 9. Consider a closed surface S. The electric field in the interor of S is zero, while on S itself the field acquires non-zero values and is directed normal to S at all points of this surface. Show that (a) S is an equipotential surface; (b) the interior of S is a space of constant potential, equal to that on S; (c) the total charge in the interor of S is zero.

• 10. By means of a simple example show that Coulomb’s law follows directly from Gauss’ law.

Problems

• 1. Consider a closed surface S inside a uniform electrostatic field $\overrightarrow{E}$. (a) Show that the total electric flux through S is zero. (b) Show that the total electric charge in the interior of S is zero.

  Solution: Let V be the volume enclosed by S. By Gauss’ law, the total flux through S is proportional to the total charge Q_in enclosed by S. Using Gauss’ integral theorem and taking into account that $\overrightarrow{E}$ is a constant vector, we have:

$$ \frac{Q_{in}}{\varepsilon_0}={\oint}_S\overrightarrow{E}\cdot \overrightarrow{da}={\int}_V\left(\overrightarrow{\nabla}\cdot \overrightarrow{E}\right)\; dv=0\operatorname{}\left( since\ \mathit{\operatorname{div}}\ \overrightarrow{E}=0\right) $$

• 2. Is it possible for an electrostatic field of the form $\overrightarrow{E}=F\left(x,y,z\right)\kern0.22em {\hat{u}}_x$ to exist? What do you conclude regarding the potential V of such a field?

  Solution: An electrostatic field must be irrotational:

$$ \overrightarrow{\nabla}\times \overrightarrow{E}=0\kern0.6em \Rightarrow \kern0.6em \left|\begin{array}{ccc}{\hat{u}}_x& {\hat{u}}_y& {\hat{u}}_z\\ {}\frac{\partial }{\partial x}& \frac{\partial }{\partial y}& \frac{\partial }{\partial z}\\ {}F& 0& 0\end{array}\right|=0\kern0.6em \Rightarrow \kern0.6em \frac{\partial F}{\partial z}\kern0.32em {\hat{u}}_y-\kern0.5em \frac{\partial F}{\partial y}\kern0.32em {\hat{u}}_z=0 $$

Now, for a vector to vanish, each component of it must be zero:

$$ \frac{\partial F}{\partial y}=\frac{\partial F}{\partial z}=0\kern0.6em \Rightarrow \kern0.6em F=F(x),\operatorname{}\mathrm{so}\ \mathrm{that}\ \overrightarrow{E}=F(x)\kern0.22em {\hat{u}}_x $$

Then, given that $\overrightarrow{E}=-\overrightarrow{\nabla}V$,

$$ \frac{\partial V}{\partial x}\kern0.22em {\hat{u}}_x+\frac{\partial V}{\partial y}\kern0.22em {\hat{u}}_y+\frac{\partial V}{\partial z}\kern0.22em {\hat{u}}_z=-F(x)\;{\hat{u}}_x $$

Equating corresponding coefficients on the two sides, we have:

$$ \frac{\partial V}{\partial x}=-F(x),\kern0.48em \frac{\partial V}{\partial y}=\frac{\partial V}{\partial z}=0\kern0.6em \Rightarrow \kern0.6em V=V(x),\kern0.48em \frac{dV}{dx}=-F(x) $$

• 3. Prove the following statements with regard to an electrostatic field: (a) the electric field lines are oriented in the direction of maximum decrease of the electric potential; (b) a positive charge that is initially at rest tends to move in the direction of decreasing potential, while a negative charge moves in the opposite way; (c) any charge (positive or negative) tends to move in the direction in which its potential energy is decreasing.

  Solution: (a) The orientation of the field lines is determined everywhere by the direction of the electric field $\overrightarrow{E}$, which is tangent to these lines. Now, for an elementary displacement $d\kern0.1em \overrightarrow{r}$ within the electric field, the corresponding change of the potential is

$$ dV=\left(\overrightarrow{\nabla}V\right)\cdot d\kern0.1em \overrightarrow{r}=-\overrightarrow{E}\cdot d\kern0.1em \overrightarrow{r} $$

In particular, for a displacement $d\kern0.1em \overrightarrow{r}$ along a field line, in the direction of orientation of the line, the element $d\kern0.1em \overrightarrow{r}$ has the direction of $\overrightarrow{E}$ and the scalar product $\overrightarrow{E}\cdot d\kern0.1em \overrightarrow{r}$ attains its maximum value. The change dV of the potential thus admits an absolutely maximum negative value. That is, the decrease of the potential is greatest along a field line, in the direction of orientation of the latter.

(b) The force on the charge is $\overrightarrow{F}=q\kern0.1em \overrightarrow{E}$. If q>0, the force is in the direction of $\overrightarrow{E}$, thus in the direction of maximum decrease of V. This will therefore be the direction of motion of a positive charge that is initially at rest (a negative charge will move in the opposite direction, i.e., that of increasing V).

(c) For an elementary displacement $d\kern0.1em \overrightarrow{r}$ within the electric field, the change of potential energy of a charge q is

$$ dU=\left(\overrightarrow{\nabla}U\right)\cdot d\kern0.1em \overrightarrow{r}=-\overrightarrow{F}\cdot d\kern0.1em \overrightarrow{r} $$

If q is initially at rest, the displacement $d\kern0.1em \overrightarrow{r}$ will be in the direction of the force $\overrightarrow{F}$ and the scalar product $\overrightarrow{F}\cdot d\kern0.1em \overrightarrow{r}$ will attain a maximum value. Thus the change dU of the potential energy of q will admit an absolutely maximum negative value. The charge q will therefore move in the direction of maximum decrease of its potential energy, regardless of the sign of q!

• 4. Two charged conductors Α and Β carry net charges +Q and –Q , respectively. Show that the electric potential of Α is greater than that of Β.

  Solution: We recall that the electric potential assumes a constant value at all points occupied by a conductor (whether on its surface or in its interior). Let V_A and V_B be the potentials of the two conductors. Along an arbitrary path connecting Α with Β,

$$ {V}_A-{V}_B={\int}_A^B\overrightarrow{E}\cdot \overrightarrow{d\kern0em l} $$

(1)

where $\overrightarrow{E}$ is the electric field along this path. Without loss of generality (given that the value of the integral is independent of the choice of path) we may assume that we move from Α to Β along an electric field line. Such a line is always oriented from the positively charged conductor Α to the negatively charged conductor Β, in accordance with the orientation of the field $\overrightarrow{E}$ (can you justify this?). Along the chosen path the vectors $\overrightarrow{E}$ and $\overrightarrow{d\kern0em l}$ are in the same direction, so that $\overrightarrow{E}\cdot \overrightarrow{d\kern0em l}>0$. The integral in (1) thus assumes a positive value and, therefore, V_A – V_B >0 ⇒ V_A > V_B.

• 5. In the interior of a conductor there is a cavity that contains no charges (Fig. 5.9). The conductor is assumed to be in electrostatic equilibrium. (a) Show that the electric field within the cavity is zero. (b) Show that the total charge on the surface of the cavity is zero. (c) We now place a charge Q inside the cavity. Find the total charge induced on the wall of the cavity, as well as the total charge on the surface of the conductor.

  Solution: (a) Since the cavity contains no charges, there can be no electric field lines beginning or ending inside the cavity. Also, since the conductor is in electrostatic equilibrium, the charge density in its interior is zero everywhere, which means that there are no nonzero charges there as well. Thus there are no field lines beginning or ending inside the conductor either. Any field line must therefore begin and end on the wall of the cavity, directed from a positive to a negative charge.

Fig. 5.9

A conductor with a cavity in its interior

We consider a closed path C such that a section of it lies inside the cavity and coincides with an electric field line. The electric field $\overrightarrow{E}$ inside the cavity is thus tangential at every point of that section of C. Taking into account that $\overrightarrow{E}=0$ in the interior of the conductor, we have:

$$ {\oint}_C\overrightarrow{E}\cdot \overrightarrow{dl}={\int}_{cavity}\overrightarrow{E}\cdot \overrightarrow{dl}+{\int}_{conductor}\overrightarrow{E}\cdot \overrightarrow{dl}={\int}_{cavity}\mid \overrightarrow{E}\mid \mid \overrightarrow{dl}\mid \kern0.48em >0 $$

which cannot be correct, given that $\oint \overrightarrow{E}\cdot \overrightarrow{dl}=0$ for any electrostatic field. The assumption we made, that there is a nonzero electric field inside the cavity, was therefore wrong. That is, we must have $\overrightarrow{E}=0$ inside the cavity.

• (b) We consider a closed surface S inside the conductor, surrounding the cavity. At every point of S we have that $\overrightarrow{E}=0$. Let Q be the total charge on the surface of the cavity. The surface S encloses no other charges, given that the charge density both in the interior of the conductor and in the interior of the cavity is zero. By Gauss’ law and by taking into account that Q_in=Q and that $\overrightarrow{E}=0$ on S, we have:

$$ {\oint}_S\overrightarrow{E}\cdot \overrightarrow{da}=0=\frac{Q}{\varepsilon_0}\kern0.5em \Rightarrow \kern0.5em Q=0\ on\ the\ surface\ of\ the\ cavity $$

Note that the above results are valid even if there exists a nonzero electric field in the exterior of the conductor. That is, the cavity is electrically isolated from the outside world, being “protected”, so to speak, by the conductor surrounding it.

(c) We consider again a closed surface S inside the conductor, surrounding the cavity. At every point of S we have that $\overrightarrow{E}=0$, as before. Let Q_wall be the total induced charge on the wall of the cavity. The total charge enclosed by S is Q_in=Q+Q_wall. By Gauss’ law and by taking into account that $\overrightarrow{E}=0$ on S, we have:

$$ {\oint}_S\overrightarrow{E}\cdot \overrightarrow{da}=0=\frac{Q_{in}}{\varepsilon_0}\kern0.5em \Rightarrow \kern0.5em {Q}_{in}=Q+{Q}_{wall}=0\kern0.5em \Rightarrow \kern0.5em {Q}_{wall}=-Q $$

Now, the conductor is electrically neutral and there is no net electric charge in its interior. So, since the wall of the cavity carries a charge –Q, there must necessarily be a charge +Q on the surface of the conductor. This surface charge makes it known to us that there is a charge Q inside the cavity!

• 6. A solid metal sphere of radius R carries a total positive charge Q uniformly distributed over its surface (Fig. 5.10). Determine the electric potential both inside and outside the sphere. (Assume that V=0 at an infinite distance from the sphere.)

Fig. 5.10

A metal sphere of radius R, carrying a total positive charge Q on its surface

Solution: By the spherical symmetry of the problem the electric field $\overrightarrow{E}$ outside the sphere is radial and directed outward (since Q>0), while its magnitude |$\overrightarrow{E}$|=E is constant over any spherical surface concentric with the sphere. Hence the external field is of the form

$$ \overrightarrow{E}=E(r)\;\hat{r},\kern0.5em r>R $$

(1)

where r is the distance from the center of the sphere. Since the solid sphere is conducting, it will be $\overrightarrow{E}=0$ in its interior, i.e., for r<R. To find the electric field in the exterior of the sphere (r>R) we apply Gauss’ law on a spherical surface S of radius r>R, concentric with the sphere. The total charge enclosed by S is Q_in=Q:

$$ {\oint}_S\overrightarrow{E}\cdot \overrightarrow{da}=\frac{Q_{in}}{\varepsilon_0}=\frac{Q}{\varepsilon_0}\kern0.5em \mathrm{where}\kern0.5em \overrightarrow{E}\cdot \overrightarrow{da}=\left[E(r)\;\hat{r}\right]\cdot \left[(da)\kern0.2em \hat{r}\right]=E(r)\; da\kern0.6em \Rightarrow $$

$$ \frac{Q}{\varepsilon_0}={\oint}_SE(r)\; da=E(r)\;{\oint}_S da=E(r)\;\left(4\pi {r}^2\right)\kern0.36em \Rightarrow \kern0.36em $$

$$ E(r)=\frac{1}{4{\pi \varepsilon}_0}\kern0.24em \frac{Q}{r^2}\kern0.36em ,\kern0.6em r>R $$

(2)

From (1) and (2) we see that the electric field in the exterior of the sphere is the same as the Coulomb field of a hypothetical point charge Q placed at the center of the sphere! As can be shown, relation (2) is valid for r=R also, thus yielding the electric field on the surface of the sphere (note that the field is noncontinuous at r=R).

We now seek the electric potential V(r). In general, for a displacement from position a to position b within the electric field, we have:

$$ \kern0.22em {\int}_a^b\overrightarrow{E}\cdot d\kern0.1em \overrightarrow{r}={V}_a-{V}_b, $$

which is independent of the path joining the end points a and b. Now,

$$ \overrightarrow{E}\cdot d\kern0.1em \overrightarrow{r}=E(r)\;\hat{r}\cdot d\kern0.1em \overrightarrow{r}=\frac{E(r)}{r}\kern0.24em \overrightarrow{r}\cdot d\kern0.1em \overrightarrow{r}=\frac{E(r)}{r}\kern0.24em r\kern0.1em dr=E(r)\; dr $$

where use has been made of Eq. (5.13). Hence,

$$ {V}_a-{V}_b={\int}_a^bE(r)\; dr $$

(3)

where E(r) is given by relation (2) for r>R, while E(r)=0 for r<R. For the potential in the exterior of the sphere, we choose point a to be on a spherical surface of radius r>R where the potential is V(r) (equipotential surface), while point b is assumed to be at infinity (r=∞) where V_∞=0 :

$$ V(r)-{V}_{\infty }={\int}_r^{\infty }E(r)\; dr=\frac{Q}{4{\pi \varepsilon}_0}\kern0.24em {\int}_r^{\infty}\frac{dr}{r^2}=-\kern0.3em \frac{Q}{4{\pi \varepsilon}_0}\kern0.24em \left(0-\frac{1}{r}\right)\kern0.48em \Rightarrow $$

$$ V(r)=\frac{1}{4{\pi \varepsilon}_0}\kern0.24em \frac{Q}{r}\kern0.36em ,\kern0.6em r\ge R $$

Notice again that the potential in the exterior of the sphere is the same as the Coulomb potential due to a point charge Q placed at the center of the sphere. For the potential in the interior of the sphere, we choose point a to be on a spherical surface of radius r<R where the potential is V(r); we take point b on the surface of the sphere where V(R)=Q/4πε₀R ; and, we use (3) by taking into account that E(r)=0 for r<R :

$$ V(r)-V(R)={\int}_r^RE(r)\; dr=0\kern0.6em \Rightarrow \kern0.6em $$

$$ V(r)=V(R)=\frac{Q}{4{\pi \varepsilon}_0R}\kern0.36em ,\kern0.6em r\le R $$

We notice that the space occupied by the sphere is a space of constant potential.

Exercise: Show that, if in place of the solid sphere we had a uniformly charged spherical shell, our results would be exactly the same! (To evaluate the electric field in the empty space bounded by the shell, use again Gauss’ law, observing that there are no electric charges in this region.)

• 7. A spherical capacitor consists of an inner conducting sphere of radius a and charge +Q, surrounded by a concentric conducting spherical shell of radius b and charge – Q (Fig. 5.11). (a) Evaluate the capacitance of the system. (b) Show that the electric field in the exterior of the capacitor (r>b) is zero and determine the electric potential both inside (a<r<b) and outside (r>b) the capacitor (by r we denote the distance from the center of the spheres).

  Solution: (a) Due to the spherical symmetry of the problem, the electric field in the interior of the capacitor (a<r<b) is of the form

Fig. 5.11

A spherical capacitor carrying a charge ±Q

$$ \overrightarrow{E}=E(r)\;\hat{r} $$

(1)

That is, the field is radial and directed from the inner to the outer sphere, while its magnitude |$\overrightarrow{E}$|=E has a constant value over any spherical surface concentric with the two conducting spheres.

We apply Gauss’ law for a spherical surface S of radius r, noting that the total charge enclosed by S is Q_in=+Q :

$$ {\oint}_S\overrightarrow{E}\cdot \overrightarrow{da}=\frac{Q_{in}}{\varepsilon_0}\Rightarrow \kern0.36em \frac{Q}{\varepsilon_0}={\oint}_S\left[E(r)\;\hat{r}\right]\cdot \left[(da)\;\hat{r}\right]={\oint}_SE(r)\; da=E(r)\;{\oint}_S da=E(r)\;\left(4\pi {r}^2\right)\Rightarrow E(r)=\frac{1}{4{\pi \varepsilon}_0}\kern0.24em \frac{Q}{r^2}\kern0.36em ,\kern0.6em a<r<b $$

(2)

The surfaces of the two spheres are equipotentials. To find the potential difference V=V_a – V_b between the spheres, we consider an arbitrary path from the inner to the outer sphere:

$$ V={V}_a-{V}_b={\int}_a^b\overrightarrow{E}\cdot d\kern0.1em \overrightarrow{r}={\int}_a^bE(r)\; dr $$

(3)

where we have used the relation $\overrightarrow{E}\cdot d\kern0.1em \overrightarrow{r}=E(r)\; dr$ (see Prob. 6). Substituting (2) into (3), we have:

$$ V=\frac{Q}{4{\pi \varepsilon}_0}\kern0.24em {\int}_a^b\frac{dr}{r^2}=\frac{Q}{4{\pi \varepsilon}_0}\kern0.36em \left(\frac{1}{a}-\frac{1}{b}\right)=\frac{Q}{4{\pi \varepsilon}_0}\kern0.24em \frac{\left(b-a\right)}{a\kern0.1em b} $$

The capacitance is then $C=\frac{Q}{V}=4{\pi \varepsilon}_0\kern0.22em \frac{a\kern0.1em b}{\left(b-a\right)}$.

• (b) In the exterior of the capacitor (r>b) the electric field will again be of the form (1). We apply Gauss’ law for a spherical surface S of radius r>b, noting that the total charge enclosed by S is now Q_in= Q + (- Q)=0 :

$$ {\oint}_S\overrightarrow{E}\cdot \overrightarrow{da}=\frac{Q_{in}}{\varepsilon_0}=0\kern0.36em \Rightarrow \kern0.36em E(r)\;\left(4\pi {r}^2\right)=0\kern0.36em \Rightarrow \kern0.36em E(r)=0 $$

Thus $\overrightarrow{E}=0$ for r>b. Let now V(r) be the potential on a spherical equipotential surface of radius r>b. For an arbitrary path connecting this surface to the outer conducting sphere, we have:

$$ V(r)-{V}_b={\int}_r^b\overrightarrow{E}\cdot d\kern0.1em \overrightarrow{r}=0\kern0.48em \Rightarrow \kern0.48em V(r)={V}_b= const. $$

By arbitrarily assuming that V_b=0, we have that V(r)=0 for r>b. In the interior of the capacitor (a<r<b) the electric field is given by (1) and (2). In this case we have (remembering that V_b=0):

$$ V(r)-{V}_b={\int}_r^b\overrightarrow{E}\cdot d\kern0.1em \overrightarrow{r}={\int}_r^bE(r)\; dr=\frac{Q}{4{\pi \varepsilon}_0}\kern0.24em {\int}_r^b\frac{dr}{r^2}\kern0.6em \Rightarrow $$

$$ V(r)=\frac{Q}{4{\pi \varepsilon}_0}\kern0.24em \left(\frac{1}{r}-\frac{1}{b}\right)\kern0.36em ,\kern0.6em a\le r\le b $$