Applying Gauss's Law

We apply Gauss's law to find the electric field due to a given charge distribution and we can also find the charge distribution from a given electric field if the enclosing surface is symmetric so that the integral $\Phi {\rm{ }} = \oint {E\cos \theta dA} = \frac{q}{{{\epsilon_0}}}$ can be evaluated easily. In this article we find the electric field due to various charge distributions using Gauss's law.

Charges on Conductors

Before diving into the applications of Gauss law you should know that the charge always lies on the outer surface of a conductor. When excess charge is added to a conductor the charge always lies at rest on the outer surface of the conductor. This can be explained in terms of the electrostatic situation of the charge.

In electrostatic situation the charge remains at rest not in motion. Since the excess charge added to the conductor is at rest, there shouldn't be any electric field inside the conductor, and if there is any electric field inside the conductor the charge will move which disturbs the electrostatic situation. So, there should not be any electric field inside the conductor, and no electric field means no charge. Which concludes that if there is no charge inside the conductor, there is no electric field which disturbs the electrostatic situation and this is valid only if the charge lies on the outer surface of the conductor.

Figure 1 The charge lies on the outer surface of a conductor.

Let's check this with Gauss's Law. We make a Gaussian surface exactly like the one shown in Figure 1. For the electrostatic situation there shouldn't be any electric field inside the conductor, so there shouldn't be any electric field inside the Gaussian surface which means there does not exist any charge inside the Gaussian surface. Therefore, there is no charge inside the Gaussian surface and it means the charge should lie on the conductor's outer surface.

Electric Field of a Uniformly Charged Conducting Sphere

Consider a uniformly charged conducting sphere (Figure 2) of radius $r$ and charge $q$. We calculate the electric field produced by the charge on the sphere at various points inside or outside the sphere. First we calculate the electric field due to the charge distribution on the sphere outside the sphere and therefore enclose the sphere of radius $r$ by a concentric Gaussian sphere of radius $R$.

Figure 2 The Gaussian sphere inside the sphere does not enclose any charge and therefore the electric field inside the sphere is zero.

In each case we create a Gaussian surface, the point where we are calculating the electric field always lies on the Gaussian surface. Note that the sphere is symmetric and the electric field is radially outward at each point of the sphere. And we know from Gauss's law (applying Gauss's law),

\[{\rm{ }}\Phi = \oint {E\cos \theta dA} {\rm{ }} = \frac{q}{{{\epsilon_0}}}{\rm{ }}\]

Since the sphere is symmetric and the electric field is radially outward the electric field at every point on the Gaussian surface is uniform and perpendicular to the area element $dA$. And $\cos \theta =\cos 0=1$. So,

\[\begin{align*} \oint {EdA} {\rm{ }} &= \frac{q}{{{\epsilon_0}}}\\ {\rm{or, }}\quad E\oint {dA} &= \frac{q}{{{\epsilon_0}}}\\ {\rm{or,}}\quad {\rm{ }}EA &= \frac{q}{{{\epsilon_0}}}\\ {\rm{or,}}\quad E(4\pi {R^2}) &= \frac{q}{{{\epsilon_0}}}\\ \therefore E &= \frac{q}{{4\pi {\epsilon_0}{R^2}}} = k\frac{q}{{{R^2}}} \tag{1} \label{1} \end{align*}\]

In this way we can determine the electric field at any distance $R$ form the centre of the charged sphere. Our result shows that the electric field outside the sphere is the same as if all the charge were concentrated at the centre of the sphere, that is the electric field is the same as that of a point charge at the centre of the sphere.

Now we determine the electric field inside the charged sphere and in this case we make a Gaussian sphere of radius $R'$ inside the sphere. In electrostatic situation i.e. charges not in motion the electric field inside the charged sphere is zero. The Gaussian surface which is inside the conductor ($R' < r$) encloses no charge and electric field inside the conductor is zero.

At the surface of the charged sphere, the Gaussian sphere has radius $r$, the same as the radius of the charged sphere. Therefore the electric field is:

\[E = \frac{q}{{4\pi {\epsilon_0}{r^2}}}{\rm{ = }}k\frac{q}{{{r^2}}} \tag{2} \label{2}\]

The electric field at the surface of the charged sphere is also the same as though all the charge were concentrated at the centre.

We recently determined the electric field of an uniformly charged conducting sphere. Now we determine the same thing for a uniformly charged insulating sphere where the charge is distributed uniformly throughout its volume.

Electric Field of a Uniformly Charged Insulating Sphere

Consider an insulating sphere of radius $r$ with net charge $q$ distributed uniformly throughout its volume in Fig:11.18. Now we determine the electric field due to that charge distribution at various points inside, on the surface and outside the sphere.

Figure 3 In case of an insulating uniformly charged sphere the Gaussian surface inside the sphere encloses some charge and therefore has electric field.

First we attempt to find the electric field inside the sphere and therefore make a concentric Gaussian sphere of radius $R'$. Here $\rho $ is the volume charge density which is the total charge divided by the volume of the sphere. So $\rho $ is,

\[\rho =\frac{q}{\tfrac{4}{3}\pi {{r}^{3}}}=\frac{3q}{4\pi {{r}^{3}}} \nonumber\]

Now the charge inside the Gaussian surface inside the sphere($r > R'$) is the volume of the Gaussian surface ${\textstyle{4 \over 3}}\pi {{R'}^3}$ multiplied by the volume charge density. So the new charge enclosed by the Gaussian surface q' is,

\[q' = \left( {\frac{{3q}}{{4\pi {r^3}}}} \right)\left( {\frac{4}{3}\pi {{R'}^3}} \right) = q\frac{{{{R'}^3}}}{{{r^3}}}{\rm{ }}\]

And the electric flux through the Gaussian surface inside the sphere is:

\[\Phi = \oint {E\cos \theta dA} = \frac{{q'}}{{{\epsilon_0}}}\]

The sphere is symmetric and charge is distributed uniformly throughout its volume so the electric field is radially outward and also uniform at every point on the Gaussian surface. And you know $\cos \theta =\cos 0=1$,

\[\begin{align*} \oint {EdA} {\rm{ }} &= \frac{{q'}}{{{\epsilon_0}}}\\ {\rm{or, }}\quad E\oint {dA} &= \frac{{q'}}{{{\epsilon_0}}}\\ {\rm{or,}}\quad {\rm{ }}EA &= \frac{{q'}}{{{\epsilon_0}}}\\ {\rm{or,}}\quad E(4\pi R{'^2}) &= \frac{{q\frac{{R{'^3}}}{{{r^3}}}}}{{{\epsilon_0}}}\\ \therefore E &= \frac{{qR{'^3}}}{{4\pi {\epsilon_0}R{'^2}{r^3}}} = k\frac{{qR'}}{{{r^3}}} \tag{3} \end{align*}\]

This is the expression for the electric field at every point on the Gaussian surface inside the sphere at a distance $R'$ form the centre. At surface of the sphere $R' = r$ and the electric field is,

\[E = k\frac{q}{{{r^2}}} \tag{4}\]

This is the same expression as that of conducting sphere in the previous application of Gauss's law. And the electric field is the same as if all the charge were concentrated at the centre of the sphere i.e. the charge on the sphere behaves like a point charge at the centre of the sphere.

To determine the electric field outside the sphere we again make a Gaussian sphere of radius $R$($R>r$) outside the sphere. Note that the Gaussian sphere is concentric with the original sphere. In this case the charge enclosed by the Gaussian surface is $q$ and we can use Gauss's law to calculate the electric field at any point on the Gaussian surface outside the sphere. So, Gauss's law gives

\[\begin{align*} E(4\pi {R^2}) &= \frac{q}{{{\epsilon_0}}}\\ \therefore E &= \frac{q}{{4\pi {\epsilon_0}{R^2}}} = k\frac{q}{{{R^2}}} \tag{5} \end{align*}\]

If you noticed the electric fields inside, on the surface and outside the sphere, you'll find that the electric field increases as $R'$ increases from $R'$ to $r$ inside the sphere and decreases as the distance increases outside the sphere ($R > r$).

Electric Field of a Line of Charge

In this case we consider that the charge $q$ is distributed uniformly in a line and forms a line of charge. Here we find the electric field at a perpendicular distance from the line of charge. We apply symmetry considerations and use Gauss's law to find the electric field. When you make a Gaussian surface to solve problems using Gauss's law you can make any kind of Gaussian surface either regular or irregular but a trick is that we make the Gaussian surface symmetrical with the charge distribution so that we can easily evaluate the Gauss's law equation (see Figure 4). Here the line of charge has cylindrical symmetry, so we apply the same cylindrical symmetry in our Gaussian surface.

Figure 4 The line of charge has cylindrical symmetry, so we use a cylinder as the Gaussian surface. The electric field is perpendicularly outward from the line of charge.

We first determine the electric flux through each ends of the cylinder and then through the curved surface. Finally we get the total electric flux by adding them together. The electric field of the line of charge is radially outward. It is because there is no such thing as the component of electric field parallel to the line of charge or tangent to the curved Gaussian surface or any other component and can not be concluded that the electric field is not radially outward. The charge is distributed uniformly throughout the line or wire, the electric field is uniform.

Since the electric field is radially outward, it is parallel to the ends of the Gaussian cylinder and hence the electric flux through the ends of the cylinder is zero. The electric field due to the line of charge is perpendicular to the curved surface. The total charge enclosed by the Gaussian surface is the liner charge density (charge per unit length) $\lambda $ multiplied by the length of the Gaussian cylinder $l$. If the Gaussian cylinder has radius $r$, the area of the curved surface of the Gaussian surface is $2\pi rl$. Now the electric field can be determined by using Gauss's law,

\[\begin{align*} E(2\pi rl) &= \frac{{\lambda l}}{{{\epsilon_0}}}\\ {\rm{or, }}\quad E &= \frac{\lambda }{{2\pi {\epsilon_0}r}} = k\frac{{2\lambda }}{r} \tag{6} \end{align*}\]

This expression is the same as the expression of the electric field of an infinite length line of charge we obtained in the electric field calculation without using Gauss's law. This is because we have considered a small portion of the line of charge in our Gaussian cylinder where all the electric field lines are radially outward which satisfies the condition of being the perpendicular distance $r$ is small enough in comparison to the length of the line of charge (also satisfies the point where we are calculating the Electric field (Gaussian surface) is close enough to the line of charge).

Electric Field of an Infinite Plane Sheet of Charge

Consider an infinite plane sheet of charge. The infinite plane sheet of charge is an idealization which works only if the point where the electric field to be calculated is close enough to the sheet compared to the sheet's dimensions and not too near the edges. The charge is distributed uniformly throughout the sheet and produces the electric field radially outward on either side of the sheet. If there is any component of the electric field parallel to the sheet, then we need to explain why the electric field has parallel component. So, the electric field is perpendicularly outward from the sheet. Now we make a Gaussian surface, a cylinder with its ends each having area $A$. The cylinder's ends are parallel to the sheet and the sheet is at the middle point of the axis of the cylinder. See Figure 5.

Figure 5 The sheet produces electric field radially outward perpendicular to the ends of the Gaussian cylinder but parallel to the curved surface.

Let the sheet has total charge $q$ and the surface charge density i.e. charge per unit area be $\sigma $. Now we divide the Gaussian surface into different parts and find the electric flux through each of those parts and obtain the total flux by adding them. But the electric field is parallel to the curved surface and perpendicular to the ends of the cylinder. Therefore the electric flux through the curved Gaussian surface is zero. Let the cylinder has radius $r$ and the surface charge density of the sheet is $\sigma$. The charge enclosed by the Gaussian surface is $\sigma A$ (surface charge density multiplied by the sheet area enclosed by the Gaussian surface). Now the electric flux through the Gaussian surface is

\[\begin{align*} EA + EA + 0 &= \frac{{\sigma A}}{{{\epsilon_0}}}\\ \therefore E &= \frac{\sigma }{{2{\epsilon_0}}} \tag{7} \end{align*}\]

Here 0 the electric flux is the electric flux through the curved Gaussian surface and $EA$ through each ends of the Gaussian cylinder. The electric field is independent of the distance from the sheet so the electric field is uniform and perpendicular to the sheet.

Electric Field between Two Charged Parallel Plates

Consider two oppositely charged conducting plates parallel to each other and we are going to find the electric field between those plates as shown in Figure 6. Note that both plates have the same surface charge density $\sigma$, that is charge per unit area. There is electrostatic attraction between the charge on opposite faces of the plates and the electric field has a direction from positive face towards negative face, so the electric field is zero on the left side of the negatively charged plate and on the right side of the positively charged plate.

Figure 6 Electric field between oppositely charged parallel plates.

Note that the charges are accumulated on the opposite faces of the plates , that is the charges are accumulated at one face of each plate. Now take a look at the Gaussian surface $G_1$ which is a cylinder where the electric field is parallel to the curved surface and perpendicular to its left end. There is no electric field to the right end of $G_1$ because the electric field due to positively charged plate $P_2$ is equal and opposite to the electric field due to negatively charged plate $P_1$. Note that for an infinite plane sheet of charge the electric field is independent of the distance from the sheet as we obtained in the previous application of Gauss's law. Therefore, the electric flux through the right end and the curved surface of $G_1$ is zero. If $A$ is the area of one of the ends, the total electric flux through the Gaussian surface $G_1$ which encloses the charge $\sigma A$ is

\[\begin{align*} 0 + 0 + EA = \frac{{\sigma A}}{{{\epsilon_0}}}\\ \therefore E = \frac{\sigma }{{{\epsilon_0}}} \tag{8} \end{align*}\]

You can also obtain the same result by making a Gaussian surface ${{G}_{2}}$ on the negatively charged plate. You can also make a Gaussian surface $G_3$ as shown in Figure 6 to make sure that the electric field is zero on the right side of the positively charged plate. The Gaussian surface $G_3$ does not enclose any charge because the charges are accumulated on the opposite faces of the plates due to electrostatic interaction, so the electric field on the right side of the positively charged plate is zero. The electric field is also zero on the left side of the negatively charged plate due to the same reason as that with the positively charged plate.

Electromagnetism

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