# Electric Field

Category : Electrostatic

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Electric Field
Electric Flux

# Electric Field

Electrostatic is one of the main subjects in any physics courses of high schools and colleges. The concept of electric field, which is a topic of electrostatic, are taught in a simple, elegant and concise way in this section with some typical problems with descriptive answer. You can solve more problems in the to strengthen your understanding and get better grade in your final or midterm exams.

Let us consider an small charged sphere of charge $q$, as shown in the figure, located at point $A$. When another particle of charge $q^{'}$ brought near the charge $q$ the force $\vec F$ exerts on it by $q$. In this case, we can say that the charge $q$ has given a property to its surrounding which causes a force acts on the charge $q^{'}$. In the other words, the space around the charge $q$ has formed an electric field that whose effect perceive by the force acting on $q^{'}$.

To find that whether there is an electric field at a certain point or not, we must place a tiny positive charged particle, called test charge, at that point (why tiny? Since this charge has its own electric field and if its electric field is not small then conflict with the principal electric field of the problem). If there is a force acting on the test charge consequently we have an electric field at that point. The electric field $\vec E$ at a certain point is defined as the ratio of the electric force $\vec F$ to the unit charge $q_0$ experienced by a charge at that point: $\vec E=\frac{\vec F}{q_0}$

Force is a vector quantity so electric field is also a vector quantity. In SI units, in which the unit of force is $\mathrm N$ and the unit of electric charge is $\mathrm C$, the unit of electric field is $\mathrm {N/C}$.

Note $\mathrm 1$: In electrostatic literature, the location of point charge is called the source point and the point where the electric field or force are calculating called field point.

## Visualization of electric field

The concept of electric field is somewhat confusing because of electric field cannot be realized directly by senses. To illustrate electric fields, one can use the notion of electric field lines. These lines have the following properties:

1. The field lines at each point have the same direction as the electric force acting on the positive charge at that point.
2. These lines for a positive charge always points away from the charge but towards the negative charge.
3. The field line at each point shows the direction of electric field at that point and the electric field at each point is tangent to the field line passing through that point and is in the same direction of it.
4. In each region which the electric field is stronger the field lines are closer and denser.
5. These lines cannot intersect each other that is from each point in space only one field line passes. In the other words, at each point of space there is only one electric field which is the net electric field.
6. Electric field lines start from positive charges and end on negative charges.

In the below the electric field around some charged objects has been shown.

In the space between two charged parallel plate of equal magnitude and opposite sign the electric field lines are parallel to each other. In the other words, at each point within the space between the plates the direction and magnitude of the electric field is constant. This field is called uniform electric field.

### Conceptual problems

1) Define electric force and explain types of it.
The force which two charged particle exerts on each other called the electric force.

2) Define Coulomb’s law and write its relation.
The magnitude of the electric force (repulsive or attractive) between the two charge $q_1$ and $q_2$ which is apart from each other by $r$ is directly proportional to product of $|q_1 q_2|$ and is inversely proportional to the distance squared $r^2$. One can summarize these proportionalities into the following relation called Coulomb’s law as
$F=k\frac{|q_1q_2|}{r^2}$

3) Define electric dipole and draw the corresponding field lines of its electric field.
A configuration consists of two equal but opposite charges at distance $r$ is called electric dipole.

## Electric field due to point charge

To obtain the electric field of a point charge $q$ at distance $r$ from it, as shown in the figure, we must place another charged particle $q_0$ at that point. According to the Coulomb’s law the force $F$ acting on the $q_0$ is
$F=\frac{1}{4\pi\epsilon_0}\frac{|q_1q_2|}{r^2}$

By using the definition of electric field, we have $E=\frac{F}{q_0} \quad \Longrightarrow \quad E=\frac{1}{4\pi\epsilon_0}\frac{q}{r^2}$

This relation shows that the magnitude of electric field due to a point charge is directly proportional to the magnitude of charge and inversely proportional to distance squared from it.

## Electric field due to a collection of point charges

If we have many point charges in a region of space, at each point of space there is a unique electric field. This field is the resultant of the fields that are produced by each charge in the absence of other electric charges at that point. This prescription is called the Superposition Principle. In a few words,
The superposition principle states that the electric field or force due to a collection of point charges on a specific charge is the vector sum of each field or force due to each individual charges that exert on that charge. $\vec E_{net}=\vec E_1+\vec E_2+\dots+\vec E_n \quad \text{or} \quad \vec F_{net}=\vec F_1+\vec F_2+\dots+\vec F_n$

1) The electric field due to charges $q_1=2\, \mathrm {\mu C}$ and $q_2=32\, \mathrm {\mu C}$ at distance $16\, \mathrm {cm}$ from charge $q_2$ is zero. What is the distance between the two charges?

Solution:

Since the two charges $q_1$ and $q_2$ are positive, somewhere between them the net electric field must be zero that is at the desired point the magnitude of fields is equal (remember that the electric field of a positive charge at field point is outward). Therefore, we get $\vec E_{net}=0 \quad \Longrightarrow \quad |\vec E_1|=|\vec E_2|$

Now use the definition of electric field to evaluate the relation above:

$k\frac{|q_1|}{r_1^2}=k\frac{|q_2|}{r^{2}} \quad \Longrightarrow \quad \frac{2}{x^{2}}=\frac{32}{16^{2}}$

Taking square root of both sides, we obtain

$\frac{1}{x}=\frac{4}{16} \quad \Longrightarrow \quad x=4 \mathrm {cm}$

As shown in the figure, the distance of two charges is $d=x+16=4+16=20\, \mathrm {cm}$.

2) In the figure, three  equal charges $q_1=q_2=q_3=+4\, \mathrm {\mu C}$ are located on the perimeter of a sphere of diameter $12\, \mathrm {cm}$. Find the net electric field, in terms of unit vectors $\hat i,\hat j$ at the center of sphere.

Solution:
The electric fields of charges $q_1$ and $q_3$ at the center of sphere are equal in magnitude and opposite in direction. Since the magnitude of charges are the same $q_1=q_3$ and are located at equal distance from the center so using definition of electric field we have
$E_1=k\frac{|q_1|}{r^2}=E_3 \quad , \quad \vec E_1=-\vec E_3$
Therefore, the resultant of field vectors at point $\mathrm O$ is, using superposition principle of fields, equal to the field $\vec E_2$.
$\vec E_{net,O}=\underbrace{\vec E_1+\vec E_3}_{0}+\vec E_2=\vec E_2$
But the electric field $\vec E_2$ lies in forth quadrant along the radius of sphere which makes angle of $53^\circ$ relative to the  $+x$ axis as shown in the figure. As mentioned already, in such cases we must decompose the vector into its components in $x$ and $y$ directions.

First use definition of electric field of a point charge to find the magnitude of $\vec E_2$ as
$\vec E_2=k\frac{|q_2|}{r^{2}}=\left(9\times 10^{9}\, \mathrm {N.\frac{m^2}{C^2}}\right)\frac{4\times 10^{-6}\, \mathrm C}{\left(\frac{12}{2}\times 10^{-2}\,\mathrm{cm}\right)^2}=10^{7}\, \mathrm{\frac{N}{C}}$
From elementary geometry we can decompose the vector $\vec E_2$ as follows
$\vec E_2=\underbrace{|\vec E_2|\,\cos 53^\circ}_{E_{2x}}\left(-\hat i\right)+\underbrace{|\vec E_2|\,\sin 53^\circ}_{E_{2y}}\left(\hat j\right)=10^{7}(0.6)\left(\hat i\right)+10^{7}(0.8)\left(-\hat j\right)$
By factoring and rearranging above relation, we get
$\vec E_2=10^{7}\left(-0.6 \hat i+0.8 \hat j\right)=6 \times 10^{6} \hat i-8\times 10^{6} \hat j$

3) As shown in the figure, the two charges $q_1$ and $q_2$ are fixed at the corners of lower side of a isosceles triangle. If the electric field vector at point $A$ (in SI) is $\vec E_A=\left(7.2 \times 10^{4}\right)\hat i$, determine the type and magnitude of electric charges $q_1$ and $q_2$.

Solution:
solution is straightforward. Use the superposition principle and find two relations between the magnitude of charges. Therefore,
$\vec E_A=\vec E_1+\vec E_2$
$\vec E_A=k \frac{|q_1|}{r_1^2}\hat r_1+k \frac{|q_2|}{r_2^2}\hat r_2$
Where in the second equality we have used the electric field of a point particle. $r$s are the unit vectors in arbitrary direction (since we have no knowledge about being positive or negative of charges) that must be found as we proceed. By decomposing unit vectors in $x$ and $y$ directions, and noting that in isosceles triangle $r_1=r_2=d$, we have

\begin{align*}
\vec E_A&=\frac{k}{d^{2}}{|q_1|\left(\cos \alpha\, \hat i+\sin \alpha\, \hat j\right)+|q_2|\left(\cos \alpha \left(-\hat i\right)+\sin \alpha\, \hat j\right)}\\
&=\frac{k}{d^2}{\cos \alpha\, \left(|q_1|-|q_2|\right)\hat i+\sin \alpha\, \left(|q_1|+|q_2|\right)\hat j}
\end{align*}
To decompose the unit vectors we have assumed the charges are positive. Because of electric field at point $A$ is in $x$ direction, so the $j$ component of the right hand side of above must be vanishes and its $i$ components must be equal to the left part as
\begin{gather*}
\vec E_A=\frac{k}{d^2}\,{\cos \alpha \left(|q_1|-|q_2|\right)\hat i+\underbrace{\sin \alpha \left(|q_1|+|q_2|\right)}_{0}\hat j}\\
|q_1|+|q_2|=0\\
7.2\times 10^{4}=\frac{k}{d^2}\, \cos \alpha\, \left(|q_1|-|q_2|\right)
\end{gather*}

The first relation says that the magnitude of charges is opposite each other i.e. $|q_1|=-|q_2|$. From the second equation one can find the \begin{gather*} 2|q_1|\frac{k}{d^2}\,\cos \alpha =7.2\times 10^{4}\\ \Longrightarrow |q_1|=3.6\times 10^{4} \frac{d^2}{k\,\cos \alpha} \end{gather*}

Use the Pythagorean theorem in the left triangle to find the value of $\cos \alpha$. \begin{gather*} 10^2=6^2+x^2 \quad \Longrightarrow \quad x=8\, \rm {cm}\\ \cos \alpha =\frac{8}{10} \end{gather*}

By substituting the given data, we obtain $|q_1|=3.6\times 10^4 \frac{\left(10\times 10^{-2} \rm {m}\right)^{2}}{\left(9\times 10^{9}\right)\left(0.8\right)}=0.5\times 10^{-9}\, \rm {C}=0.5\, \rm {nC}$

Thus, $|q_1|=-|q_2|=0.5\, \rm {nC}$. Our initial assumption that the charges is positive does not correct. Therefore, we must choose correctly one of them to be positive and the other negative. By choosing $q_1$ to be positive and $q_2$ negative, one can arrive at the right net electric field at point $A$.

Electric field, Field line, uniform electric field, electric dipole, Superposition Principle