What is an Electric Field?
Definition of Electric Field
Electric Field Intensity
The intensity of an electric field at a point is defined as the force experienced per unit positive test charge placed at that point. It is represented by the symbol E.
Formula of Electric Field Intensity
The electric field intensity is given by:
`\vec E = \vec F/q_0`
SI Unit and Dimensional Formula of Electric Field
SI Unit:
Newton per Coulomb `N/C` or Volt per meter `V/m`.
Dimensional Formula:
`[M^1 L^1 T^{-3} A^{-1}]`
Where:
E = Electric field intensity
F = Force acting on the test charge
`q_0 =` Magnitude of the positive test charge
Force on a Charge in an Electric Field
Formula:
`\vec F = q_0 \vec E`
This equation shows that the force acting on a charge is directly proportional to both the magnitude of the charge and the strength of the electric field.
- Force `\prop`q
- Force `\prop ` E
Important Points:
- If the charge is positive, the force acts in the same direction as the electric field.
- If the charge is negative, the force acts in the opposite direction of the electric field.
Key Concept of Electric Field
`\star` The electric field intensity is a vector quantity, meaning it has both magnitude and direction.
`\star` Its SI unit is Newton per Coulomb (N/C). It can also be expressed as Volt per meter (V/m).
`\star` The dimensional formula of electric field intensity is: `[M^1L^1T^{-3}A^{-1}]`.
Quick Revision:
- An electric field is a vector quantity.
- SI unit: N/C or V/m
- Dimension: `[M^1 L^1 T^{- 3} A^{-1}]`.
- The direction of the electric field is the same as the direction of the force on a positive test charge.
Mathematically, electric field intensity is expressed as the ratio of the force to the unit charge.
`\vec E = \frac{\vec F}{q_0}`
If the particle of charge q is placed in `\vec E` electric field intensity, then the force acting on charge q is
`\vec F = q \vec E`
Note -
- If a particle has a positive charge, the force acting on it is in the same direction as the electric field.
- If a particle has a negative charge, the force acting on it is in the opposite direction to the electric field.
- For a positive point charge, the electric field is directed radially outward (away from the charge).
- For a negative point charge, the electric field is directed radially inward (towards the charge).
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| Electric Field Lines digram |
Types of Electric Field
Uniform electric field
Characteristics:
Non-uniform electric field
Characteristics:
Variable electric field
Characteristics:
- Depends on time (E changes with time).
- Important in electromagnetic Waves and AC circuits.
- If electric field lines are straight, parallel, and equally spaced, it shows a uniform field.
- If field lines are not straight or not equally spaced, it shows a Non-uniform field.
- Closer field lines show a strong electric field.
- Farther field lines show a weak electric field.
- Electric field lines never intersect each other.
Electric Field Due to a Point Charge Derivation
Suppose a charge +q is located at point O. Consider a point P at a distance r from O, where a test charge `q_0` is placed.
The electric field at a point is defined as the force experienced per unit positive test charge.
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| Electric Field due to Point Charge |
From Coulomb's law, the force on the test charge is:
`\vec F = \frac{K q q_0}{r^2} \hat r`
By the definition of the electric field
`\vec E = \frac{\vec F}{q_0} `
`\vec E = \frac{K q q_0}{q_0 r^2}\hat r`
`\vec E = \frac{K q }{ r^2}\hat r`
Important Points
`\star` The direction of the electric field is along `\hat r`, which represents the radial direction. It is outward for a positive charge and inward for a negative charge (radially outward for +q, inward for -q).
`\star` The electric field is inversely proportional to the square of the distance (`E = 1/r^2`).
`\star` The electric field depends only on the source charge q, not on the test charge `q_0`.
`\star` The graph between electric field intensity and distance is -
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| The graph between electric field intensity and distance |
Do you know
- Depends only on source charge q.
- Independent of the test charge.
- Follows the inverse square law.
Electric Field in a Medium
`\vec E = \frac{K q }{ r^2}\hat r`
`\vec E = \frac{1}{4 \pi \epsilon_0} \frac{ q }{ r^2}\hat r`
` E = \frac{1}{4 \pi \epsilon_0 \epsilon_r} \frac{ q }{ r^2}\hat r`
` E_{medium} = \frac{E_{vacu um}}{\epsilon_r}`
Thus, for a dielectric medium, the value of electric field intensity is `\epsilon_r` times less than that in vacuum.
Electric Field due to a System of Charges
The net electric field at a point is the vector sum of electric fields due to all charges.
The electric field intensity at a point P is
`\vec E_p = \vec E_1 + \vec E_2 + ..............+ \vec E_n`
`\vec E_1 = \frac{K q_1}{r_1^2} \hat{r}_1`
`\vec E_2 = \frac{K q_2}{r_2^2} \hat{r}_2`
`\vec E_3 = \frac{K q_3}{r_3^2} \hat{r}_3`
Test Your Knowledge: Electric Field and Electric Field Lines MCQs
Frequently Asked Questions (FAQs)-Electric Field and Electric Field Lines
Answer: According to the principle of superposition, the electric field intensity at a point is equal to the vector sum of the different intensities due to all the charges.
2. How do you calculate the electric field due to a system of charges?
Answer: According to the principle of superposition of charges, to find the electric field at a point, the electric fields generated due to all the charges are calculated and their vector sum is taken. This vector sum is equal to the intensity of the electric field at that point.
3. Can the electric field at a point be zero due to a system of charges?
Answer: Yes, the electric field intensity due to a group of charges at a point can be zero; this happens when the vector sum of the intensities due to all the charges at that point is zero.
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