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Category : Electrostatic

All topics about electrostatic in physics suitable for high school and colleges are collected here with a brief course note and tens of solved problems in other sections. 


Electric field : definition and tutorial
Electric Flux: Definition & Solved Examples
Electric Field - Problems and Solution
Coulomb's Law: Solved Problems for High School and College
Electric flux: Problems with Solutions for AP Physics

Electric flux: Problems with Solutions for AP Physics

Electric flux problems with detailed solutions are presented for uniform and non-uniform electric fields. Each solution is a self-tutorial so that the definition of electric flux and its formula are provided. 

Electric flux of uniform electric fields:

Problem (1): A uniform electric field with a magnitude of $E=400\,{\rm N/C}$ incident on a plane surface of area $A=10\,{\rm m^2}$ and makes an angle of $\theta=30^\circ$ with it. Find the electric flux through this surface?

Solution: electric flux is defined as the amount of electric field passing through a surface of area $A$ with formula \[\Phi_e=\vec{E} \cdot \vec{A}=E\,A\,\cos\theta \] where dot ($\cdot$) is the dot product between electric field and area vector and $\theta$ is the angle between $\vec{E}$ and the normal vector to the plane.

In this problem, the angle between electric field and normal to the plane is $60^\circ$ so we get \begin{align*} \Phi_e &=EA\,\cos \theta\\&=400\times 10\times \underbrace{\cos 30^\circ}_{1/2}\\&=2000\quad {\rm N.m^2/C}\end{align*}


Problem (2): What is the magnitude of electric flux of a constant E of $4\,{\rm N/C}$ in the z-direction through a rectangle with surface area $4\,{\rm m^2}$ in the xy-plane?

Solution: To compute electric flux, we need the magnitude of the electric field, area of surface, and the angle between E and normal to the surface. Here, $\vec{E}=4\,\hat{k}\,{\rm N/C}$ and area $A=4\,{\rm m^2}$ are given explicitly, but the angle isn't. 

We know that the xy-plane is directed toward up or down with unit normal $\hat{n}=(0,0,\hat{k})$ or $\hat{n'}=(0,0,-\hat{k})$ respectively. Thus, by choosing $\hat{n}$ the angle between E and Area is $\theta=0$ and the flux is given as \begin{align*} \Phi_e&=EA\,\cos \theta\\&=4\times 4\times \underbrace{\cos 0^\circ}_{1}\\&=16\quad {\rm N.m^2/C}\end{align*}


Problem (3): A cube of side 1 cm is placed within an electric field of 100 N/C oriented in the x-direction. What is the electric flux through each of the surfaces of the cube and then find the net flux? 

Solution: In this problem, $E$ and area of each surface $A=L^2=(0.01)^2=10^{-4} \,{\rm m^2}$ are the constant. To find the electric flux through each surface of a cube, first, determine the normal vector (a vector perpendicular to the surface and pointing outward) to each surface by inspection. 

For simplicity, we denote the area of each surface with $A_{ij}$, where $ij$ are the sides of the surface. With this introduction, we have
\begin{align*} 
\Phi_{xy} &=EA_{xy}\cos 90^\circ \\&=100\times 10^{-4} \times 0\\&=0\\
\Phi_{yz-I} &=EA_{yz}\cos 180^\circ \\&=100\times 10^{-4} \times -1\\&=-10^{-2} \,{\rm N.m^2/C}\\
\Phi_{xz} &=EA_{xz}\cos 90^\circ \\&=100\times 10^{-4} \times 0\\&=0\\
\Phi'_{xy} &=EA_{xy}\cos 90^\circ \\&=100\times 10^{-4} \times 0\\&=0\\
\Phi_{yz-II} &=EA_{yz}\cos 0^\circ \\&=100\times 10^{-4} \times 1\\&=10^{-2} \,{\rm N.m^2/C}\\
\Phi'_{xz} &=EA_{xz}\cos 90^\circ \\&=100\times 10^{-4} \times 0\\&=0 \end{align*}
The net flux is determined by summing all above fluxes as \[\Phi_{net}=0\]


Problem (4): A $2\,{\rm cm} \times 2\,{\rm cm}$ square lies in the xy-plane. Find the electric flux through the square for each of the following electric field vectors?
(a) $\vec{E}=(50\,\hat{i}+20\,\hat{j})\,{\rm N/C}$
(b) $\vec{E}=(50\,\hat{k}+20\,\hat{j})\,{\rm N/C}$

Solution: the electric flux of a uniform electric field through a flat surface is defined as the scalar product of electric field and area vector $\vec{A}=A\hat{n}$, where $\hat{n}$ is a vector perpendicular to the surface and points outward. 

The normal vector to a surface on the xy-plane is parallel to z-axis i.e. $\hat{n}=(0,0,1)$.

Recall that the scalar product of two vectors $\vec{A}=A_x \hat{i}+A_y \hat{j}$ and $\vec{B}=B_x \hat{i}+B_y \hat{j}$ is given as $\vec{A}\cdot \vec{B}=A_x B_x+A_y B_y$.

(a) Therefore, the electric flux through a flat surface on the xy-plane is
\begin{align*} \Phi_e &=\vec{E}\cdot \vec{A} \\&=(50\,\hat{i}+20\,\hat{j})\cdot (+\hat{k})\\&=50\,\underbrace{\hat{i}\cdot\hat{j}}_{0}+20\,\underbrace{\hat{j}\cdot\hat{k}}_{0}\\&=0 \end{align*}

(b) Again, we have \begin{align*} \Phi_e &=\vec{E}\cdot \vec{A} \\&=(50\,\hat{k}+20\,\hat{j})\cdot (+\hat{k})\\&=50\,\underbrace{\hat{k}\cdot\hat{k}}_{1}+20\,\underbrace{\hat{j}\cdot\hat{k}}_{0}\\&=50\quad {\rm N.m^2/C} \end{align*}


Problem (5): A circle of radius $3\,{\rm cm}$ lies in the yz-plane and in the path of an electric field of $\vec{E}=(100\hat{i}+200\hat{j}-50\hat{k})\,{\rm N/C}$. Find the electric flux through the circle?

Solution: A vector perpendicular and in outward or inward direction to the yz-plane is parallel to the x-axis i.e. $\hat{n}=(1,0,0)$ or $\hat{n'}=(-1,0,0)$ so we choose the area vector is $\vec{A}=A\,\hat{n}$ where $A=\pi r^2$ is the area of the circle. 

Note: the normal vector on an open surface can point in either direction. In this case, we must choose one of them.  

By definition of electric flux as the dot product of $\vec{E}$ and area vector $\vec{E}$, we obtain 
\begin{align*} \Phi_e &=\vec{E}\cdot \vec{A} \\&=(100\,\hat{i}+200\,\hat{j}-50\,\hat{k})\cdot (A\,\hat{i})\\&=100\,A\,(\underbrace{\hat{i}\cdot\hat{i}}_{1})+200\,A\,(\underbrace{\hat{j}\cdot\hat{i}}_{0})-50\,A\,(\underbrace{\hat{k} \cdot \hat{i}}_{0})\\&=100\,\pi\,(0.03)^2\\&=28.27\quad {\rm N.m^2/C} \end{align*}
In above the area of the circle is found as $A=\pi r^2=\pi\times (0.03)^2$.


Problem (6): A flat surface with an area of 20 squared meters lies in the xz-plane where a uniform electric field of $\vec{E}=5\hat{i}+4\hat{j}+5\hat{k}$ exists. Find the electric flux through this surface? 

Solution: the vector perpendicular (normal vector) to the xz-plane is parallel to the y-axis so we choose $\hat{n}=(0,1,0)$. Definition of electric flux says that the scalar product of electric field $\vec{E}$ and area vector $\vec{A}=A\hat{n}$ gives the amount of electric field line passing through a surface, therefore we have \begin{align*} \Phi_e &=\vec{E}\cdot \vec{A}\\&=(5\,\hat{i}+4\,\hat{j}+5\hat{k})\cdot (A\,\hat{j})\\&=4A\\&=4(20)=80\quad {\rm N.m^2/C} \end{align*} In above we used the definition of scalar product. 


Problem (7): Find the electric flux through the surface with sides of $15\,{\rm cm}\times 15\,{\rm cm}$ shown in the figure below.

electric radiates at an angle to the surface

Solution: first find the angle between the electric field and the vector perpendicular to the plane. The normal vector to the plane is shown as upward. 

In this problem, the electric field makes an angle of $30^\circ$ with the plane. To find the angle with the normal first coincide the tails of two vectors and then determine the required angle which is $\theta=90^\circ+30^\circ=120^\circ$. 

Next using the definition of electric flux, $\Phi_e=EA\,\cos \theta$, we get \begin{align*} \Phi_e&=EA\,\cos \theta \\&=150\times (0.15)^{2}\times \cos 120^\circ\\&=-1.125\quad {\rm N.m^2/C}\end{align*}.


Problem (8): A square surface with sides of $1\,{\rm m}\times 1\,{\rm m}$ located over the xy-plane, where a constant electric field with a magnitude of $200\,{\rm N.m^2/C}$ presents. The direction of the electric field vector is depicted in the figure. What is the total electric flux through the open surface? 

electric field make angle with normal vector

Solution: since the electric field is constant and surface is flat so we can use the electric flux formula $\Phi_e=EA\cos \theta$. The angle between $E$ and $A$ are shown in the figure, so the electric flux through the surfacee is \begin{align*} \Phi_e &=E\,A\,\cos \theta\\&=200\times (1)^2 \times \cos (180^\circ-72^\circ)\\&=-61.8\quad {\rm N.m^2/C}\end{align*} the negative electric flux indicates that E and normal vector are in opposite direction. 

flux through a rectangular surface from front view


Problem (9): The electric field intensity at all points in space is given by $\vec{E}=\sqrt{3}\hat{i}-\hat{j}\quad \Big({\rm \frac Vm}\Big)$. A square frame of side 1 meter is shown in the figure. The point N lies in the xy-plane. The initial angle between line ON and the x-axis is $60^\circ$. Find the magnitude of the electric flux through the area enclosed in the square frame LMNO.

flux through a rectangular plane and unifrom electric field

Solution: first determine the normal vector to the surface of LMNO. As shown in the figure, the perpendicular vector is $\hat{n}=\sin \theta (-\hat{i})+\cos \theta (+\hat{j})$ where $\theta$ is shown in the figure. 

angle between electric field and normal vector

Now using the definition of electric flux, $\Phi_e=\vec{E}\cdot\vec{A}$ where $\vec{A}=A\,\hat{n}$ is the area vector, we get \begin{align*} \Phi_e &=\vec{E}\cdot\vec{A}\\&=(\sqrt{3}\hat{i}-\hat{j})\cdot \Big(A\,\sin \theta (-\hat{i})+A\,\cos \theta (+\hat{j})\Big)\\&=-A\sqrt{3}\sin 60^\circ (\underbrace{\hat{i}\cdot \hat{i}}_{1})-A\cos 60^\circ\,(\underbrace{\hat{j}\cdot\hat{j}}_{1})\\&=-A\sqrt{3}\,\frac{\sqrt{3}}2  - A\,\frac 12\\&=-2A\\&=-2(1 \times 1)\\&=-2 \quad {\rm V.m}\end{align*} The negative value of electric flux indicates that the electric field and normal vector are in the opposite direction. 


Electric flux of non-uniform electric fields: 

Problem (10): A non-uniform electric field is given by the expression \[\vec{E}=ay\hat{i}+bz\hat{j}+cx\hat{k}\] Where $a,b,c$ are constants, $\hat{i},\hat{j},\hat{k}$ are unit vectors in the $x,y,z$ directions, respectively. Determine the electric flux through a rectangular surface in the xy-plane, extending from $x=0$ to $x=w$ and from $y=0$ to $y=h$.

Solution: The electric flux of a non-uniform electric field through any surface is defined to be in integral form as  \[{\Phi }_e=\oint_S{\vec{E}\cdot\hat{n}dA}\] Where $S$ stands for the surface we are integrating over, $\hat{n}$ is the unit vector normal to the surface and $\vec{E}$ is the electric field over the surface.

In this case, since the surface lies in the xy-plane, so unit vector normal to it is $\hat{n}=\hat{k}$ \begin{align*} {\Phi }_e&=\oint_S{\vec{E}\cdot \hat{n}\,dA}\\ &=\int{\Big(ay\hat{i}+bz\hat{j}+cx\hat{k}\Big) \cdot (dx\,dy \hat{k})} \\ &=c\,h\,{\Big(\frac{1}{2}x^2\Big)}^w_0\\ &=\frac{1}{2}chw^{2} \end{align*}


Problem (11): The cubical surface of side length $L=12\ {\rm cm}$ is shown in the electric field $\vec{E}=\left(950\,y\,\hat{i}+650\,z\,\hat{k}\right)\ {\rm V/m}$. Find the electric flux through the top face of the cube.

Flux through faces of a cube

Solution: By definition, the electric flux passing through any surface $A$ is the number of field lines penetrating it. In the mathematical form \[\Phi_e=\int_S{\vec{E}\cdot \hat{n}dA}\] Where $\hat{n}$ is the unit vector normal to the surface $A$.

In this problem, the normal vector is parallel to the z-axis that is $\hat{n}=\hat{k}$. So 
\begin{align*}
\Phi_e&=\int{\left(950\,y\, \hat{i}+650\, z\,\hat{k}\right)\cdot\hat{k}\ dA}\\
&=\int{\Big(950\, y\,(\underbrace{\hat{i}\cdot \hat{k}}_{0})+650\, z\,(\underbrace{\hat{k}\cdot \hat{k}}_{1})\Big)\,(\underbrace{dx\,dy}_{dA})}\\
&=650\,(0.12)\int{dxdy} \end{align*} Where the last integral is the area of the surface which is integrated over. Thus the total flux through the given surface is 
\begin{align*} \Phi_e &=650(0.12)L^2\\&=650\times (0.12)^3\\&=1.1232\quad {\rm N.m^2/C}\end{align*} 


Problem (12): A hemispherical shell of radius R is placed in an electric field E which is parallel to its axis. What is the flux ${\Phi }_e$ of the electric field through the shell?

electric flux through a hemisphere

Solution: 
By definition, the electric flux passing through any surface with area element $dA$ is the integral of the scalar product of the normal component of the electric field and area of the surface that is \[{\Phi }_e=\int_S{\vec{E}\cdot \hat{n}\,dA}\] Where $\hat{n}$ is the unit vector normal to the surface and $S$ is the surface over which the integral is evaluated. 
 Angle between electric field and normal to the sphere is shown.
In the case of hemisphere or sphere, the unit vector is along the radius i.e. $\hat{n}=\hat{r}$. Since $\vec{E}$ is perpendicular to the plane of the great circle of the hemisphere so the scalar product of $E$ and $\hat{r}$ is $E{\cos \theta\ }$. The range of polar angle is from $\theta=0$ to $\theta=\pi/2\ $. Therefore,
\begin{align*} {\Phi }_e&=\int^{\frac{\pi}{2}}_0{\vec{E}\cdot \hat{r}dA} \\&= \int^{2\pi}_0{d\phi}\int^{\frac{\pi}{2}}_0{E\,{\cos \theta\ }\ R^2\,{\sin \theta\ }d\theta}\\ &= 2\pi ER^2\int^{\frac{\pi}{2}}_0{\underbrace{{\sin \theta\ }{\cos \theta\ }}_{\frac{1}{2}{\sin  2\theta\ }}d\theta}\\ &=2\pi ER^2\left(\frac{1}{2}\right){\left(-\frac{1}{2}{\cos  2\theta\ }\right)}^{\frac{\pi}{2}}_0\\ &=-\frac{1}{2}\pi ER^2\left({\cos  2\frac{\pi}{2}\ }-{\cos  0\ }\right)\\&=\pi ER^2 \end{align*} In above, $dA=R^2\,{\sin \theta\ }d\theta d\phi $ is the area element of the sphere. 

Alternate Solution:
All of the electric field lines through the circle at the bottom of the hemisphere, passing through the area of hemisphere. So by calculating the flux through the circle, one can find the flux through the hemisphere but the important thing to remember is that, by convention, the flux through the circle is incoming and negative of the outgoing flux through the hemisphere that is ${\Phi }_e\left(circle\right)=-{\Phi }_e(hemisphere)$. 

Therefore, by definition of the electric flux through a surface, we get \begin{align*} {\Phi }_e \left(circle\right)&=\vec{E}\cdot \hat{n}\,A\\&=E\,(+\hat{k})\cdot (-\hat{k})\,(\pi R^2)\\&=-\pi ER^2\end{align*} Note: the normal vector $\hat{n}$ is defined always in outward direction of the surface.


Problem (13): An square of side L lies in the xy-plane. The electric field crosses the square in the z-direction as $\vec{E}=\frac {E_0}{a}\,y\,\hat{k}$. What is the electric flux through the surface as shown in the figure?

Solution: remember that the electric flux is a surface integral. Take an infinitesimal strip of width $dy$ and length $L$ so that the electric field over it is constant. Consequently, the area vector becomes $d\vec{A}=a\,dy\,\hat{k}$. Therefore, the integral form of definition of electric flux gets
\begin{align*} \Phi_e&=\int_S{\vec{E}\cdot\hat{n}\,dA}\\&=\int{(\frac {E_0}{a}\,y\,\hat{k})\cdot (a\,dy\,\hat{k})}\\&=\frac {E_0}{a}\,a\int_{0}^{a}{y\,dy}\\&=\frac {E_0}{a}\,a\,\Big(\frac 12 y^{2}\Big)_0^{a}\\&=\frac 12 E_0 a^2 \end{align*}

Note: for a non-uniform electric field, the integral definition of electric flux must be used. 


Problem (14): A rectangular flat surface is placed on the xy-plane. In this region of space, there is a non-uniform electric field of $\vec{E}=E_0 x^2 \hat{k}$, where $E_0$ is a constant. What is the electric flux through the rectangular surface? 

electric flux passes through a rectangular surface with non-uniform electric field

Solution: Since the electric field is not constant over the surface so the integral definition of electric flux must be used. 

An infinitesimal strip is shown

To use this method, we must first construct an area element such that the electric field is constant across it. In this problem, along the y-direction, $\vec{E}$ is constant so we choose an strip with area $dA=a\,dx$. Thus we get the net electric flux through this open surface as below \begin{align*} \Phi_e&=\int_S{\vec{E}\cdot \hat{n}\,dA}\\&=\int_0^b{\Big(E_0\,x^2\hat{k}\Big)\cdot \hat{k}(a\,dx)}\\&=E_0\,a\int_0^b{x^2 dx}\\&=aE_0 \Big(\frac 13 x^3\Big)_0^b\\&=\frac 13 aE_0\,b^3 \end{align*}


All of the above electric flux problems suitable for high schools and colleges.
 


Date Created: 10/24/2020

Author: PhysicsExams


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