Electric Field: Definition, Application, Solved Examples

In this world, generally, we need to have contact in order to apply some force, but this is not always the case. A charged balloon attracts the hair without even coming in contact with it. How does this happen? Is it similar to how a body experiences a force due to the gravitational field? Or how the iron nail is attracted towards the magnet? Is an electric field similar to these? What are electric field lines? What do they signify?

Let us read the article further to understand more about the electric field.

What is a Field?

For some physical phenomena or properties, we do not require direct physical contact in order to apply force. Example: When we hold two magnets, we observe a push or a pull when we bring them closer depending on the poles, but one thing is sure they exert force without even coming into physical contact. In order to understand these types of force, we use the concept of field. Example: A charge kept in space will generate a field that will signify its presence in the space, and any other charge will experience some force when kept in the field of that charge.

Electric Field

An electric charge kept in space generates an electric field in the space around it. And if any other charge is placed at a point in that field, then it will experience a force depending on the strength of the field at that point. The field is generated from the charge and extends up to infinity. Also, then we use the concept of the electric field lines in order to understand the field better.

Electric Field Lines

Electric field lines or electric lines of force is a hypothetical concept that we use to understand the concept of Electric field. Electric field lines represent the electric field pictorially in direction and magnitude. The number of electric field lines emerging or sinking in a charge is directly proportional to the magnitude of the charge.

Electric field lines can be both straight lines or curved but they cannot be a closed curves as electric field lines cannot emerge and sink at the same point as electric field lines emerge from a positive charge and sink at a negative charge. Electric field lines cannot intersect as it will give two directions for the net electric field at the point of intersection which is not possible.

Electric Field due to Point Charges

The electric field at a point due to a point charge is given using the coulomb’s law.

According to the coulombs law, the force between the two charges is given by,

\(F = \frac{{K{q_1}{q_2}}}{{{r^2}}}\)
Where,
\({q_1},{q_2}\) are the two charges.
\(K = \frac{1}{{4\pi {\varepsilon _0}}}\) is the proportionality constant.
\(r\) is the distance between the two charges.
The force on a charge in an electric field is given by,
\(\vec F = q\vec E\)
\( \Rightarrow \vec E = \frac{{\vec F}}{q}\)
\(\vec E = \frac{{Kq}}{{{r^2}}}\)
Electric field due to more than one charge is given by the vector sum of the individual charges.
\({E_{{\text{net}}}} = {\vec E_1} + {\vec E_2} + \ldots \)

Electric Field due to Continuous Bodies

For continuous bodies, an electric field can be determined by integrating the field due to differential elements of that continuous bodies, or we can use Gauss law in many of the cases.

\(\vec E = \smallint d\vec E\)

Electric field due to a uniform charged thread of finite length is given by,

The component of the electric field in the direction perpendicular to the thread is,

\({E_ \bot } = \frac{\lambda }{{4\pi {\varepsilon _0}~d}}\left|{\sin \left({{\theta _1}} \right) + \sin \left({{\theta _2}} \right)} \right|\)

The component of the electric field in the direction parallel to the thread is,

\({E_\|} = \frac{\lambda }{{4\pi {\varepsilon _0}~d}}\left|{\cos \left({{\theta _2}} \right) – \cos \left({{\theta _1}} \right)} \right|\)

Where,
\({\theta _1}\) and \({\theta _2}\) are the angle subtended by the ends of the thread at that point
\({{d}}\) is the perpendicular distance of that point from the thread.
\(\lambda \) is the charge per unit length of the thread.
If \({\theta _1} = {\theta _2},\) then the component of the electric field in the direction parallel to the thread will be zero.
Electric field due to infinite uniformly charged thread

\(\vec E = \frac{\lambda }{{2\pi {\varepsilon _0}d}}\)

In the case of an infinite charged thread, the component of the electric field in the direction parallel to the thread will be zero.

Electric field in the direction parallel to the length of the thread is cancelled out due to symmetry, and only the component of the electric field in the direction perpendicular to the thread exists.

Electric field due to semi-infinite thread

\({E_ \bot } = \frac{\lambda }{{4\pi {\varepsilon _0}~d}}\)
\({E_\|} = \frac{\lambda }{{4\pi {\varepsilon _0}d}}\)
We can get the above result by putting in the value of \({\theta _1} = \frac{\pi }{2}\) and \({\theta _2} = 0\)
Electric field due to uniformly charged circular arc at its centre

\(\vec E = \frac{{2\lambda }}{{4\pi {\varepsilon _0}R}}\sin \left({\frac{\theta }{2}} \right)\)
Where,
\(\lambda \) is the charge per unit length of the thread.
\({{R}}\) is the radius of the arc.
\(\theta \) is the angle subtended by the arc at its centre.
Electric field due to uniformly charged circular ring at its axis

\(\vec E = \frac{{qx}}{{4\pi {\varepsilon _0}{{\left({{R^2} + {x^2}} \right)}^{\frac{3}{2}}}}}\)
Where,
\(q\) is the charge on the ring.
\(x\) is the distance of the point on its axis from the centre of the ring.
Electric field due to uniformly charged disc at a point on its axis
\(\vec E = \frac{\sigma }{{2{\varepsilon _0}}}\left({1 – \cos \left( \theta \right)} \right)\)
Where,
\(\theta \) is the angle subtended by the radius of the disc on that point.
\(\sigma \) is the surface charge density of the disc.
Electric field due to charged infinite sheet
\(\frac{\sigma }{{2{\varepsilon _0}}}\)
Where,
\(\sigma \) is the surface charge density of the disc.

Applications

Van de Graaff generator

Smoke precipitator

Sample Problems

Q.1. The charge per unit length for a circular ring varies as \(\lambda = {\lambda _0}\sin \left( \theta \right).\) Then find the electric field at its centre.

Ans: Let us consider a differential element of the ring \(dq,\) which subtends an angle \(d\theta \) at the centre at any angle \(\theta \) from the axis.
We can write the electric field at the centre due to this differential element as,
Along \(x\)-axis,
\({E_x} = \int d E\cos \left( \theta \right)\)
Along \(y\)-axis,
\({E_y} = \int d E\sin \left( \theta \right)\)
Where,
\(dE = \frac{{dq}}{{4\pi {\varepsilon _0}{R^2}}}\)
Substituting the value \(dq = \lambda Rd\theta \) and \(\lambda = {\lambda _0}\sin \left( \theta \right),\) we get,
\( \Rightarrow dE = \frac{{{\lambda _0}\sin \left( \theta \right)d\theta }}{{4\pi {\varepsilon _0}R}}\)
To find the net electric field we integrate the above expression with the limit \(0\) to \(2\pi .\)
Thus, the electric field along \(x\)-axis is,
\({E_x} = \int d E\cos \left( \theta \right)\)
\( \Rightarrow \int d E\cos \left( \theta \right) = \int_0^{2\pi } {\frac{{{\lambda _0}\sin \left( \theta \right)\cos \left( \theta \right)d\theta }}{{4\pi {\varepsilon _0}R}}} \)
\( \Rightarrow {E_x} = 0\)
The electric field along \(y\)-axis,
\(\Rightarrow {E_y} = \int d E\sin \left( \theta \right) = \int_0^{2\pi } {\frac{{{\lambda _0}{{\sin }^2}\left( \theta \right)d\theta }}{{4\pi {\varepsilon _0}R}}} \)
Using, \({\sin ^2}\left(\theta \right) = \frac{{1 – \cos \left({2\theta } \right)}}{2}\)
We get,
\(\frac{{{\lambda _0}}}{{2 \times 4\pi {\varepsilon _0}R}}\mathop \smallint \nolimits_0^{2\pi } \left({1 – \cos \left({2\theta } \right)} \right)d\theta \)
\( \Rightarrow {E_y} = \frac{{{\lambda _0}}}{{4{\varepsilon _0}R}}\)

Q.2. In the given figure with uniform charge density, find the electric field at the centre?

Ans:

The electric field due to a circular arc at the centre is in the direction of the angular bisector of the angle subtended by the arc at the centre, and if the charge is positive, the direction of the electric field is away from the centre, and if it is negative, then it is towards the arc.

Since the charge density is the same and they subtend equal angles at the centre, the magnitude of the electric field due to the three arcs will be the same.

Thus, keeping the above facts in mind, if we draw the vector diagram of the three fields we get, the resultant field to be zero.

Summary

Field is defined as the physical quantity which exists in space by virtue of which force can be experienced. Example: gravitational field, electric field, magnetic field.

Electric field is the field generated due to an electric charge at rest. Any other charge in that field will experience a force due to that field.

To understand the behaviours of the electric field, we use the concept of electric field lines.

Electric field lines represent the magnitude and the direction of the electric field at any point.

We can calculate electric field using coulombs law for electrostatics for both point charges and continuously charged bodies.

Frequently Asked Questions on Electric Field

Here, we have listed some of the frequently asked questions on Electric Field:

Q.1: What is an electric field?
Ans: Electric field is a phenomenon by virtue of which any charge experiences a force when kept in that field. All-electric charges generated an electric field.

Q.2: What is electric field lines or electric lines force?
Ans: Electric field line or electric line of force is a hypothetical line that we use to represent the electric field visually, and it represents the direction and the magnitude of the electric field lines in space.

Q.3: Is the number of electric field lines generated by a charge infinite?
Ans: No, the number of electric field lines generated by a charge is not infinite.
The number of electric field lines generated by a charge of magnitude \(q\) is \(\frac{q}{{{\varepsilon _0}}}.\)

Q.4: From where do the electric field lines emerge, and where do they sink?
Ans: Electric field emerges from a positive charge and sinks in a negative charge.

Q.5: What is the expression for electric field due to a uniformly charged thread?
Ans: Electric field due to a uniform charged thread of finite length is given by, The component of the electric field in the direction perpendicular to the thread is, \({E_ \bot } = \frac{\lambda }{{4\pi {\varepsilon _0}~d}}\left|{\sin \left({{\theta _1}} \right) + \sin \left({{\theta _2}} \right)} \right|\)
The component of the electric field in the direction parallel to the thread is, \({E_\|} = \frac{\lambda }{{4\pi {\varepsilon _0}~d}}\left|{\cos \left({{\theta _2}} \right) – \cos \left({{\theta _1}} \right)} \right|\)

Q.6: What is the expression for electric field due to a uniformly charged disc?
Ans: Electric field due to uniformly charged disc at a point on its axis
\(\vec E = \frac{\sigma }{{2{\varepsilon _0}}}\left({1 – \cos \left( \theta \right)} \right)\)

Get ahead with our practice questions on electric field here

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