# Gauss's Law Problems and Solutions

## Gauss's Law Definition:

In simple words, Gauss's law states that the net number of electric field lines leaving out of any closed surface is proportional to the net electric charge $q_{in}$ inside that volume.

On the other hand, the number of electric field lines is also defined as electric flux $\Phi_E$ passing through any closed surface. Therefore, we can arrive the following formula for Gauss's law $\Phi_E=\oint{\vec{E}\cdot d\vec{A}}=\frac{q_{in}}{\epsilon_0}$

Gauss's law simplifies the calculation of electric field due to the various charge distributions with highly symmetric.

## Solved Problems on Gauss's law:

Problem (1): Find the net electric charge inside the sphere below.

Solution: in the definition of Gauss's law, net charge mean the arithmetic sum of all charges inside the desired closed surface. Thus, in the sphere, the net charge inside is $q_{in}={\rm (+1\,nC)+(-2\,nC)=-1\,nC}$

Problem (2): A point charge of $-2\,{\rm \mu C}$ is located at the center of a cube with sides $L=5\,{\rm cm}$. What is the net electric flux through the surface?

Solution: In this problem, computing electric flux through the surface of the cube using direct definition as $\Phi_E=\vec{E}\cdot \vec{A}$ is a hard and time-consuming task.

Gauss's law is an alternative to find the flux which is simply states that divide enclosed charge by $\epsilon_0$. Thus, the flux through the above cube is calculated as below \begin{align*}\Phi_E&=\frac{q_{in}}{\epsilon_0} \\ \\&=\frac{-2\times 10^{-6}}{8.85\times 10^{-12}}\\\\&=-226\times 10^{3}\quad {\rm \frac{N\cdot m^{2}}{C}}\end{align*} The minus sign indicates that the electric field lines are entering the cube.

Problem (3): A charge is located inside the cube (is not at the center)
(a) Find the net electric flux passing through the surface of the cube?
(b) Can you find the electric field on the surface of the cube using Gauss's law?

Solution: (a) As before, using Gauss's law the net electric flux is computed as the net charge inside the cube divided by $\epsilon_0$. Therefore, \begin{align*} \Phi_E&=\frac{q_{in}}{\epsilon_0}\\\\&=\frac{3\times 10^{-9}}{8.85\times 10^{-12}}\\\\&=339\quad {\rm \frac{N\cdot m^{2}}{C}}\end{align*}

(b) Gauss's law is only applicable and computable where there is high symmetry in the problems. In this example, the distance of charge to each surface of the cube is not the same. Thus, we can not find the electric field on the surfaces of the cube.

Problem (4): A particle with charge Q is placed at the center of a cube with edges of L.
(a) What is the electric flux that passes through the entire cube?
(b) Find the electric flux through one of the faces of the cube?
(c) Calculate the electric field on one of the faces of the cube?

Solution
(a) Using Gauss's law formula, $\Phi_E=q_{in}/\epsilon_0$, the electric flux passing through all surfaces of the cube is $\Phi_E=\frac{Q}{\epsilon_0}$
(b) All above electric flux passes equally through six faces of the cube. Thus, by dividing the total flux by six surfaces of a cube we can find the flux through each of them. $\phi_E=\frac{\Phi_E}{6}=\frac{Q}{6\epsilon_0}$ Note that if the charge located everywhere except the center of the cube, we can not do this work since the flux through the surface close to the charge is greater than the flux through the surface farther to the charge.

(c) We can not use Gauss's law to find the electric field on each surface of the cube, since all points of a surface are not at the same distance from the charge inside the cube, so the electric field can not be factored out of the integral.

Problem (5): A $3.5\,{\rm cm}$-radius hemisphere contains a total charge of $6.6\times 10^{-7}\, {\rm C}$. The flux through the rounded portion of the surface is $9.8\times {10}^4 \,{\rm \frac{N\cdot m^2}{C}}$. What is the flux through the flat base of the hemisphere?

Solution: The amount of electric flux  $\Phi_E$ through any closed surface and the associated enclosed total charge is related together by Gauss's law as $\Phi_E=\frac{Q_{in}}{\epsilon_0}$ Hemisphere has two surfaces, rounded and flat base thus the total electric flux through it is as \begin{align*} \underbrace{\Phi_{E,r}+\Phi_{E,b}}_{\Phi_E}&=\frac{Q_{in}}{\epsilon_0} \\ \\ 9.8\times {10}^4+\Phi_{E,b}&=\frac{6.6\times {10}^{-7}}{8.854\times {10}^{-12}}\\ \\ \Rightarrow \quad \Phi_{E,b}&=-2.34\quad {\rm \frac{N\cdot m^2}{C}} \end{align*}

Problem (6): A charge $q$ is located exactly at the center of the sphere. Find the electric field at the surface of the sphere?

Solution: We can use Gauss's law to find either net electric flux through any closed surface or electric field on the desired surface provided that there is high enough symmetry like this example.

Using Gauss's law formula the net electric flux through the surface of the sphere is $\Phi_E=\frac{q_{in}}{\epsilon_0}=\frac{q}{\epsilon_0}$ Next, use the definition of the flux to find the electric field at the sphere's surface as below \begin{align*} \Phi_E &=\oint{\vec{E}\cdot d\vec{A}}\\ \\&=\oint{E dA \cos \theta} \\ \\ &=E \cos 0^\circ \oint{dA} \\ \\&=E(4\pi R^{2})\end{align*} In the second line, we used the definition of scalar product where $\theta$ is the angle between $\vec{E}$ and a unit vector normal (perpendicular) to the surface. In the last line, integral over all surface of the sphere gets the area of a sphere.

Combining the first and second parts of the problem, we get the electric field on the surface of the sphere as below $4\pi R^{2}E=\frac{q}{\epsilon_0} \Rightarrow E=\frac{q}{4\pi\epsilon_0}$

Note: if the above charge sits anywhere except the center of the sphere, we can not find the electric field on the sphere's surface using Gauss's law.

Problem (7): Charge $Q$ is distributed uniformly throughout an insulating sphere of radius $R$. Find the magnitude of the electric field at a point $R/2$ from the center.

Solution: If we have a symmetric configuration, as in this case, use Gauss's law to find the electric field at each point of space. To do this, one must suppose a Gaussian surface at the desired point. In this problem, draw a Gaussian surface as a sphere of radius $R/2$ and proceed as follows
$\oint{\vec{E}.d\vec{A}}=\frac{Q_{in}}{{\epsilon }_0}$
Where $Q_{in}$ is the charge inside the Gaussian surface which here is sphere of radius $R/2$ and is determined by the definition of volume charge density \begin{align*} \rho &=\frac{Q}{V} \\ \\ \Rightarrow Q_{in}&=\rho V_{Gauss} \\ \\ &=\rho \left(\frac{4}{3}\pi {\left(\frac{R}{2}\right)}^3\right)\\ \\&=\frac{1}{6}\rho \pi R^3\end{align*} Substituting above into Gauss's law, we get $E\oint{dA}=\frac{1}{6}\rho \pi R^3$ The electric field is taken out of the integral since it is constant at the location of the Gaussian sphere by symmetry considerations. The closed integral gives the area of the Gaussian surface.

Therefore \begin{align*} E\left(4\pi {\left(\frac{R}{2}\right)}^2\right)&=\frac{\frac{1}{6}\rho \pi R^3}{\epsilon_0} \\ \\ \Rightarrow E&=\frac{1}{4\pi {\epsilon }_0}\frac{2}{3}\rho \pi R\end{align*} Since the density of the sphere is uniform everywhere so substitute it by the total charge induced on the original sphere of radius $R$ as $\rho =\frac{Q}{V}=\frac{Q}{\frac{4}{3}\pi R^3}$ Therefore
\begin{align*} E&=\frac{1}{4\pi {\epsilon }_0}\frac{2}{3}\frac{Q}{\frac{4}{3}\pi R^3}\pi R\\ \\ &=\frac{1}{4\pi {\epsilon }_0}\frac{Q}{{2R}^2}\\ \\ &=\frac{Q}{8\pi {\epsilon }_0R^2}\end{align*}

Problem (8) A non-uniform, but spherically symmetric, distribution of charge has a charge density of $\rho\left(r\right)=\rho_0\left(1-\frac{r}{R}\right)$ for $(r<R)$ and $\rho\left(r\right)=0$ for $(r>R)$ where $\rho_0=\frac{3Q}{\pi R^3}$ is a positive constant.

(a) Show that the total charge contained in the charge distribution is $Q$
(b) Show that the electric field in the region ($r>R$) is identical to that produced by a point charge $Q$ at $r=0$.
(c) Obtain an expression for the electric field for $r<R$ and graph the magnitude of the $E(r)$ as a function of $r$.
(d) Find the value of $r$ at which the electric field $E(r)$ is a maximum, and find the value of that maximum field.

Solution
(a) By definition, the total charge distributed in a region of space is $Q=\int{\rho\left(r\right)dV}$
\begin{align*} Q_{tot}&=\int{\rho\left(r\right)dV}\\ &=\int^R_0{\rho_0\left(1-\frac{r}{R}\right)\ 4\pi r^2dr}\\ &= 4\pi \rho_0{\left(\frac{1}{3}r^3-\frac{r^4}{4R}\right)}^R_0\\ &=\frac{1}{3}\pi \rho_0R^3\\ &=\frac{1}{3}\pi R^3\left(\frac{3Q}{\pi R^3}\right)=Q\\ \quad \Rightarrow Q_{tot}&=Q \end{align*}
(b) Use Gauss's law to find the electric field of a charge distribution inside and outside of the desired volume. The charge configuration has spherical symmetry so by symmetry consideration, the electric field must be radial. \begin{align*} \oint{\vec{E}\cdot \hat{n}dA}&=\frac{Q_{in}}{\epsilon_0} \\ \\ E_r\oint{dA}&=\frac{Q_{in}}{\epsilon_0} \\ \\  E_r\left(4\pi r^2\right)&=\frac{Q_{in}}{\epsilon_0} \\ \\ \Rightarrow E_r&=\frac{Q_{in}}{4\pi\epsilon_0r^2}\end{align*} if $r>R$ then the net charge inside the Gaussian surface $Q_{in}=\int{\rho\left(r\right)dV}=Q$ because the Gaussian surface encloses the overall sphere. Therefore $E_r\left(r>R\right)=\frac{Q}{4\pi\epsilon_0r^2}$
(c) From part (b) , $E_r=\frac{Q_{ins}}{4\pi\epsilon_0r^2}$ but the amount of enclosed charge in a Gaussian surface inside of the sphere is  \begin{align*} Q_{in} &=\int{\rho\left(r\right)dV}\\ \\ &=\int^r_0{\rho_0\left(1-\frac{r}{R}\right)4\pi r^2dr}\\ \\ &=4\pi\epsilon_0{\left(\frac{1}{3}r^3-\frac{r^4}{4R}\right)}^r_0\\ \\ &=4\pi\epsilon_0\left(\frac{1}{3}r^3-\frac{r^4}{4R}\right) \end{align*} Therefore, the electric field in the region $r<R$ is obtained as below $\Rightarrow E_r\left(r<R\right)=\frac{\rho_0}{\epsilon_0}\left(\frac{1}{3}r-\frac{r^2}{4R}\right)$
(d) we must find the maximum value of the electric field inside the sphere so
\begin{align*}\frac{d}{dr}E_r\left(r<R\right)&=0 \\ \\ \frac{\rho_0}{\epsilon_0}\left(\frac{1}{3}-\frac{r}{2R}\right)&=0 \\ \\ \Rightarrow r&=\frac{2}{3}R\end{align*} Substituting this value of $r$ into the above electric field inside the sphere, we have
$E_r\left(r=\frac{2}{3}R\right)=\frac{\rho_0}{\epsilon_0}\left(\frac{2}{9}R-\frac{R}{9}\right)=\frac{\rho_0}{9\epsilon_0}R$

Problem (9): A solid conducting sphere carrying charge $q$ has radius $a$. It is inside a concentric hollow conducting sphere with inner radius $b$ and outer radius $c$. The hollow sphere has no net charge.
Derive expressions for the electric field magnitude in terms of the distance $r$ from the center for the region $r<a\ ,\ a<r<b\ ,\ b<r<c\$and $r>c$.
(a) Graph the magnitude of the electric field as a function of $r$ from $r=0$ to $r=2c$.
(b) What is the charge on the inner surface and on the outer surface of the hollow sphere?

Solution
In such problems, one can use Gauss's law to find the electric field everywhere.

(a) Gauss's law states that the electric flux through any closed surface $S$ is equal to the charged enclosed by it divided by $\epsilon_0$ with formula $\oint_s{\vec{E}.\hat{n}dA}=\frac{Q_{enc}}{\epsilon_0}$
To use Gauss's law, we must first consider a closed surface which is called a Gaussian surface. This surface has the same symmetry as the electric field. In this case, the Gaussian surface must be a spherical surface of radius $r$ concentric with the conducting charged sphere of radius $a$.

Note: although Gauss's law is true for any surface surrounding a charged configuration, it is useful only when we choose a Gaussian surface to match the original symmetry of the problem.

To find the electric field in the region $r<a$, the Gaussian surface is inside the sphere of radius $a$. Since inside this surface there is not any charge, so there is no flux through it, thus $\oint_s{\vec{E}\cdot \hat{n}dA}=\frac{Q_{in}}{\epsilon_0}=0\Rightarrow E\left(r<a\right)=0$ In the region $a<r<b$, the Gaussian surface encloses the charged sphere $q$, so \begin{align*}\oint_s{\vec{E}\cdot \hat{n}dA}&=\frac{Q_{in}}{\epsilon_0}\\ \\ E\oint_s{dA}&=\frac{q}{\epsilon_0} \\ \\ \Rightarrow E\left(a<r<b\right)&=\frac{q}{4\pi \epsilon_0r^2}\end{align*} Since $E$ is constant and perpendicular everywhere on the Gaussian surface, so we have taken it out of the integral.

The closed integral $\oint_s{dA}$ is the surface area of the sphere. In the case of sphere, the normal vector $\hat{n}$ is along the radial direction i.e. $\hat{n}=\hat{r}$.

Since inside the Gaussian surface, there is a positive charge so the electric field point away from the center or is radially outward i.e. $\vec{E}=E\left(r\right)\hat{r}$.

Region $b<r<c$ lies inside the conductor. Using this fact that electric field inside a conductor is zero , we get $E\left(b<r<c\right)=0$.

In region $r>c$, the net charge encloses by the Gaussian surface is $+q$, so in this region the electric field is \begin{align*}\oint_s{\vec{E}\cdot \hat{n}dA}=\frac{Q_{in}}{\epsilon_0}=\frac{q}{\epsilon_0}\\ \\ \Rightarrow E\left(r>c\right)&=\frac{q}{4\pi \epsilon_0r^2}\end{align*} The graph is as follows

(b) The free charge $q$ inside the sphere of radius $a$ induces the charge $-q$ on the inner surface and subsequently, this charge also induces the charge $+q$ on the outer surface of the spherical shell.

Page Published: 2/28/2021

Author: Ali Nemati