Step-by-Step Derivation and Real-Life Applications
When a glass rod is rubbed with a piece of silk, the rod acquires the property of attracting objects like bits of paper, etc towards it.
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This happens due to the discharge of electric charges by rubbing of insulating surfaces.
Electric charge is a property that accompanies fundamental particles, wherever they exist.
When an electric charge q₀ is held in the vicinity of another charge Q, q₀ either experience a force of attraction or repulsion.
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We say that this force is set up due to the electric field around the charge Q.
Therefore, we can say that the electric field of charge Q as space by virtue of which the presence of charge Q modifies the space around itself leading to the generation of force F on any charge q₀ held in this space, given by:
\[F = \frac{k|Q|q_|}{r^{2}}\]
Here, from the above figure, we have the following parameters
r = The separation between source charge and test charge
Q = source charge,
q1 = test charge, and
\[k = \frac{1}{4\pi \epsilon_{0}} = 9\times 10^{9} N m^{2} C^{-1}\]
The electric field due to a given electric charge Q is defined as the space around the charge in which electrostatic force of attraction or repulsion due to the charge Q can be experienced by another charge q.
Electric Field Due to a Point Charge Formula
The concept of the field was firstly introduced by Faraday.
The electric field intensity at any point is the strength of the electric field at that point.
It is defined as the force experienced by a unit positive charge placed at a particular point.
Here, if force acting on this unit positive charge +q₀ at a point r, then electric field intensity is given by:
\[\overrightarrow{E}({r}) = \frac {\overrightarrow{F}{(r)}}{q_o}\]
Hence, E is a vector quantity and is in the direction of the force and along the direction in which the test charge +q tends to move.
Its unit is \[\frac {N} {C}\].
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The electric field for +q₀ is directed radially outwards from the charge while for - q₀, it will be radially directed inwards.
Electric Field Due to a Point Charge Example
Suppose we have to calculate the electric field intensity or strength at any point P due to a point charge Q at O.
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From Fig.1 we have the following,
Here, OP = r.
Now, consider a small positive charge q₀ at P.
According to Coulomb’s law, the force of interaction between the charges q₀ and Q at P is,
\[F = \frac{1}{4\pi \epsilon_{0}} \frac{Qq_{0}}{r^{2}}\]
Where r is a unit vector directed from Q towards q₀.
We know,
\[\overrightarrow{E}({r}) = \frac { \overrightarrow{F}(r)} {q_o}\]
Therefore,
\[\overrightarrow{E} = \frac{1}{4\pi \epsilon_{0}} / r^2 (r)\]
Derivation of Electric Field Due to a Point Charge
Suppose the point charge +Q is located at A, where OA = r1.
To calculate the electric field intensity (E) at B, where OB = r2.
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From fig.2, we have:
According to Coulomb's law, the force on a small test charge q2 at B is,
\[F = \frac{1}{4 \pi \epsilon_{0}} \frac{q_{1}q_{2}(r_{12})}{r_{12^2}}\]
\[\frac{1}{4 \pi \epsilon_{0}} \frac{q_{1}q_{2}(r_{12})}{r_{12^3}}\]
\[\overrightarrow{F} = \frac{1} {4\pi \epsilon_{0}}{\frac{ q_{1}q_{2}}{|\overrightarrow{r_{2}} - \overrightarrow{r_{1}}|^{3} . (\overrightarrow{r_{2}} - \overrightarrow{r_{1}})}}\]
Here, \[AB = \overrightarrow{r_{12}} = \overrightarrow{r_{2}} - \overrightarrow{r_{1}}\]
As, \[\overrightarrow{E} = \frac{\overrightarrow{F}}{q_{2}}\]
Therefore,
\[\overrightarrow{F} = \frac{1} {4 \pi \epsilon_{0}}{\frac{ q_{1}}{|\overrightarrow{r_{2}} - \overrightarrow{r_{1}}|^{3} . (\overrightarrow{r_{2}} - \overrightarrow{r_{1}})}}\]
Hence, E is produced along AB.
Electric Field Due to a System of Point Charges
The electric field intensity at any point due to a system or group of charges is equal to the vector sum of electric field intensities due to individual charges at the same point.
Now, we would do the vector sum of electric field intensities:
\[\overrightarrow{E} = \overrightarrow{E_{1}} + \overrightarrow{E_{2}} + \overrightarrow{E_{3}} + ... + \overrightarrow{E_{n}}\]
\[\overrightarrow{E} = \frac{1}{4 \pi \epsilon_{0}} \sum_{i=1}^{i=n} \frac{\widehat{Q_{i}}}{r_{i}^{2}} . (r_{i})\]
Here,
\[r_{i}\] is the distance of the point P from the ith charge \[Q_{i}\] and \[r_{i}\] is a unit vector directed from \[\widehat{Q_{i}}\] to the point P.
ri is a unit vector directed from Qi to the point P.
Let’s say charge Q1, Q2...Qn are placed in vacuum at positions r₁, r₂,....,rₙ respectively.
The net forces at P are the vector sum of forces due to individual charges, given by,
\[\overrightarrow{F} = \frac{1}{4\pi \epsilon_{0}} q_{0} \sum_{i=1}^{i=n} \frac{\overrightarrow Q_{i}}{|\overrightarrow{r} - \overrightarrow{r_{i}}|^{3} . |\overrightarrow{r} - \overrightarrow{r_{i}}|}\]]
As, \[\overrightarrow{E} = \frac{\overrightarrow{F}}{q_{0}}\]
Therefore,
\[\overrightarrow{E} = \frac{1}{4\pi \epsilon_{0}} \sum_{i=1}^{i=n} \frac{\overrightarrow Q_{i}}{|\overrightarrow{r} - \overrightarrow{r_{i}}|^{3} . |\overrightarrow{r} - \overrightarrow{r_{i}}|}\]]
Putting \[\frac {1}{4 \pi \epsilon_{0}}\] = k
\[\overrightarrow{E} = k \frac {Q_{1}} {r_{1^2}} + k \frac {Q_{2}}{r_{2^2}} + . . . + k \frac {Q_{n}} {r_{n^2}}\]
Hence, we obtained a formula for the electric field due to a system of point charges.
FAQs on Electric Field Due to a Point Charge: Explained
1. What is the definition of electric field due to a point charge as per the CBSE 2025–26 Physics syllabus?
The electric field due to a point charge is defined as the region around a charged particle where another charge experiences an electrostatic force. Its magnitude at a distance r from a point charge Q is given by E = (1 / 4πε₀) · (Q / r²), with direction always pointing away from positive and towards negative charges.
2. How is the direction of the electric field affected by the sign of the point charge?
The direction of the electric field is radially outwards from a positive point charge and radially inwards towards a negative point charge. This direction indicates how a unit positive test charge would move in the field.
3. How can you calculate the net electric field at a point due to multiple point charges?
To find the net electric field at a point due to several point charges, calculate the electric field due to each charge at that point and then perform a vector sum of all these individual fields. Mathematically: 𝐄₍ₙₑₜ₎ = 𝐄₁ + 𝐄₂ + 𝐄₃ + ... + 𝐄ₙ.
4. Why do electric field lines never intersect each other?
Electric field lines never cross because at any single point in space, the electric field has only one unique direction. If lines crossed, it would imply two directions for the field at a point, which is not possible physically.
5. Can electric fields exist in a vacuum, and what does this imply about force between charges in space?
Yes, electric fields exist in a vacuum. Charges placed in vacuum exert forces on each other according to Coulomb’s law, meaning the electric field is a property of space around a charge, not dependent on a material medium.
6. How does the magnitude of the electric field due to a point charge change with distance?
The magnitude of the electric field due to a point charge decreases inversely with the square of the distance from the charge. Mathematically: E ∝ 1/r². Doubling the distance reduces the field strength to one-fourth.
7. Explain the significance of a unit positive test charge in calculating electric field intensity.
The unit positive test charge is used to define electric field intensity without affecting the existing field. The field strength at any point is the force experienced by this unit positive test charge placed at that point.
8. What common misconceptions do students have about the electric field and force on charges?
A common misconception is that electric field and force are the same. In fact, the electric field is a property of space created by charges, while force is experienced by a charge placed within the field. Also, some think that field lines represent paths of charges, but they only show direction and strength of the field.
9. How did Faraday contribute to the concept of an electric field due to a point charge?
Michael Faraday introduced the concept of an electric field as an invisible field surrounding electric charges, visualized using field lines, to explain how forces act at a distance even without direct contact.
10. In what way is the formula for electric field due to a point charge modified in a medium other than vacuum?
When a point charge is placed in a medium with relative permittivity εr, the formula becomes E = (1 / 4πε₀εr) · (Q / r²). The field strength decreases as the permittivity of the medium increases, compared to its value in a vacuum.
11. What happens to the electric field between two equal and opposite point charges placed close together?
The field lines in this setup form a dipole pattern. The electric field between the charges is strong and nearly uniform in the region directly between them, while outside the region, fields from each charge tend to cancel each other at far distances.
12. How can the vector nature of electric field be represented when dealing with point charges in three-dimensional space?
The vector nature of the electric field is shown using unit vectors pointing from the source charge to the point of interest. The electric field at a position 𝐫 due to a charge at 𝐫₀ is 𝐄 = (1/4πε₀) [Q / |𝐫 - 𝐫₀|²] 𝑢̂, where 𝑢̂ is the unit vector in the direction from the source to observation point.
13. What is the SI unit of electric field, and how is it derived?
The SI unit of electric field is newton per coulomb (N/C). It is derived from the definition as force (N) per unit charge (C): 1 N/C = 1 volt/meter.
14. Why is a test charge required to be very small when determining electric field intensity?
The test charge should be infinitesimal so that it does not disturb the existing electric field created by the source charge. A large test charge could alter the field pattern being measured.
15. How does the principle of superposition apply to electric fields due to multiple point charges?
The principle of superposition states that the total electric field at a point due to several point charges is the vector sum of the fields created by each charge independently, without influencing each other’s fields.
