# 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, electric field lines are also defined as electric flux $\Phi_E$ passing through any closed surface. Therefore, we can arrive at 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 the electric field associated with the distribution of a highly symmetric charge.

In the following, some questions with answers about Gauss's law are solved numerically and qualitatively which is helpful for class 12 students and college students.

## Gauss's law: Solved Practice Problems

Problem (1): Find the net electric charge inside the sphere below. Solution: In Gauss's law definition, net charge means 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 its direct definition as $\Phi_E =\vec{E}\cdot \vec{A}$ is a hard and time-consuming task.

Gauss's law is an alternative to finding the electric flux which 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, substituting the numerical values, we have \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 when there is a 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.

Need help with your homework? Get this pdf with 500+ solved physics homework. Only for $8 or download a free pdf sample. Problem (4): A particle with charge Q is placed at the center of a cube with edges of L. (a) What is the electric flux that passes through the entire cube? (b) Find the electric flux through one of the faces of the cube? (c) Calculate the electric field on one of the faces of the cube? Solution (a) Using Gauss's law formula,$\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 the 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 a charge is located everywhere except the center of a 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_Ethrough 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*} To prepare for the AP Physics C exam, these questions on the electrostatic force are also relevant. Problem (medium): A cube with sides of\ell=8.5\,\rm cm$is positioned in a uniform electric field of strength$E=7500\,\rm N/C$so that two of the faces are parallel to the field lines. (a) What is net flux passing through the cube? (b) What is the flux through each of the cube's six faces? Solution: In this problem, a cube is placed into an external uniform electric field and we are asked to find the flux through the whole and each face of the cube. Assume this situation in your mind. Assume a cube with six vectors (of magnitude unit) perpendicular to each of its faces. These vectors are called unit normal vectors,$\hat{n}$. The flux through each face is found as below $\Phi_E=EA\cos\theta$ where$\theta$is the vector between electric field vector$\vec{E}$and each of those unit vectors perpendicular to the faces,$\hat{n}$. (a) Now insert such a cube into a uniform external electric field so that two of its faces, say up and down, are in the direction of$E. Thus, the flux passing through these two faces is measured to be \begin{align*} \phi_{up}&=EA\cos 0^\circ \\&=7500\times (0.085)^2\times 1 \\&=54.2\quad \rm N\cdot m^2/C \\\\ \phi_{down}&=EA\cos 180^\circ \\&=7500\times (0.085)^2\times (-1) \\&=-54.2\quad \rm N\cdot m^2/C \end{align*} The electric fluxes through other faces of the cube are zero because their normal vectors make an angle of90^\circ$with the external electric field$\vec{E}. (b) The net electric flux is the sum of fluxes through all faces of the cube, assuming no charge is involved in the cube. Therefore, \begin{align*} \Phi_{net}&=\phi_{up}+\phi_{down}+\cdots \\ &=54.2+(-54.2)+0 \\ &=\boxed{0} \end{align*} Problem (medium): A point chargeq$is placed near a plane with approximately infinite extension. What is the flux through such a plane due to that point charge? Solution: Since the plane is infinitely extended, it does not matter how much the electric charge is close to it. But for better visualization, assume the charge to be extremely near it. In this case, exactly half the field lines, incoming or outgoing, penetrate the infinite plane. On the other hand, the total electric flux or field lines that a charge$q$produces, by using Gauss's law, is found as $\Phi=\frac{q}{\epsilon_0}$ Therefore, the total field lines that cross the infinite plane (or flux passing through that plane) is computed as below $\phi_{plane}=\frac 12 \Phi=\frac{q}{2\epsilon_0}$ Learn more about electric charge properties with solving these problems on electric charge Problem (medium): A point charge of$0.0524\,\rm \mu C$is positioned at the center of a tetrahedron. (a) What is the total electric flux through its whole surface? (b) What is the flux through one of its faces? Solution: The electric flux$\Phi_Ethrough any closed surface is related to the charge inside it by Gauss's law $\Phi_E=\frac{Q_{in}}{\epsilon_0}$ (a) According to the definition above, the total electric flux through the four sides surface (pyramid) is calculated as below \begin{align*}\Phi_E&=\frac{0.0524\times 10^{-6}}{8.85\times 10^{-12}} \\\\ &=5921\quad \rm N\cdot m^2/C \end{align*} Recall that the electric flux represents the number of field lines passing through a given surface. In this problem, about 5921 field lines pass through the entire surface of the tetrahedron. (b) The above result was the total flux passing through the four sides of the tetrahedron. Because the charge is placed at the center of the tetrahedron, it is at an equal distance from each face. Consequently, the flux through each face is one-fourth of the total flux. $\phi_E=\frac 14 \Phi_E = 1480\,\rm N\cdot m^2/C$ Practice Problem: In the figure below, a configuration of four closed surfaces and three charges of-2Q$,$+Q$, and$-Q$is shown. What is the electric flux through each surface? Solution: Electric flux$\Phi_E$or the number of field lines passing through a given closed surface is found by Gauss's law as below $\Phi_E=\frac{Q_{enclosed}}{\epsilon_0}$ As you can see, all that matters is the presence of an electric charge and a closed surface surrounding it. There are two charges inside the orange closed surface that give a total charge of$Q_{enclosed}=-2Q+Q=-Q$. Dividing this net charge by$\epsilon_0, we get the electric flux through the orange surface $\phi_{orange}=\frac{Q_{enc}}{\epsilon_0}=\frac{-Q}{\epsilon_0}$ In a similar manner, the flux through other colored closed surfaces are computed as below \begin{align*} \phi_{blue}&=\frac{Q_{enc}}{\epsilon_0}\\ &=\frac{-Q+Q}{\epsilon_0}=\boxed{0} \\\\ \phi_{black}&=\frac{Q_{enc}}{\epsilon_0} \\&=\frac{-2Q+Q-Q}{\epsilon_0} \\ &=\boxed{-\frac{2Q}{\epsilon_0}} \end{align*} No electric flux passes through the green closed surface because it does not surround any charge. $\phi_{green}=\frac{Q_{inside}}{\epsilon_0}=\boxed{0}$ Problem (medium): An electron is placed at a distance of\ell/2$just above the center of a square of edge$\ell. Find the magnitude of the electric flux that passes through the square. Solution: If we want to use Gauss's law to find the flux for such a configuration, then we should construct a closed surface around the given charge. On the other hand, recall that Gauss's law has limitations and is only applicable to situations with highly symmetric shapes or charge configurations. We know that if an electric charge (say, electron or proton) is inserted at the center of a cube, then the flux through each of its faces equals one-sixth of the total flux through that cube. In this case, if we form a cube with the given square so that the electron is positioned at its center, then the flux passing through our original square is computed with ease. The total electric flux through such an imaginary cube is \begin{align*} \Phi_{total}&=\frac{q_{inside}}{\epsilon_0} \\\\ &=\frac{-1.6\times 10^{-19}}{8.85\times 10^{-12}} \\\\ &=1.8\times 10^{-8}\,\rm N\cdot m^2/C \end{align*} As a result, the flux through that original square is $\phi_E=\frac 16 \Phi_{total}=3\quad \rm nN\cdot m^2/C$ wheren=10^{-9}$. Problem (6): A charge$qis 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, the integral over the whole 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 symmetrical 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 the sphere of radius$R/2and 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 radiusRas $\rho =\frac{Q}{V}=\frac{Q}{\frac{4}{3}\pi R^3}$ 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 the 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*} ifr>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 regionr<Ris 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 ofr$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 an 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$. (b) What are the charges on the inner surface and the outer surface of the hollow sphere? Solution In such problems, one can use Gauss's law to find the electric field everywhere. (a) Gauss's law states that the electric flux through any closed surface$S$is equal to the charge enclosed by it divided by$\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 valid 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$q, so \begin{align*}\oint_s{\vec{E}\cdot \hat{n}dA}&=\frac{Q_{in}}{\epsilon_0}\\ \\ E\oint_s{dA}&=\frac{q}{\epsilon_0} \\ \\ \Rightarrow E\left(a<r<b\right)&=\frac{q}{4\pi \epsilon_0r^2}\end{align*} SinceE$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 a sphere, the normal vector$\hat{n}$is along the radial direction i.e.$\hat{n}=\hat{r}$. Because inside the Gaussian surface, there is a positive charge, the electric field points 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 the fact that the 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$+q, so in this region the electric field is \begin{align*}\oint_s{\vec{E}\cdot \hat{n}dA}=\frac{Q_{in}}{\epsilon_0}=\frac{q}{\epsilon_0}\\ \\ \Rightarrow E\left(r>c\right)&=\frac{q}{4\pi \epsilon_0r^2}\end{align*} The graph is as follows (b) The free chargeq$inside the sphere of radius$a$induces charge$-q$on the inner surface, and subsequently, this charge also induces charge$+q\$ on the outer surface of the spherical shell.

Page Published: 2/28/2021

Last Update: Sep 16, 2022

Author: Dr. Ali Nemati