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Superposition principle states that when a number of charges are interacting with each other, then the net electrostatic force exerted on a particular charge is the vector sum of all the forces exerted on it by all other charges.
- The principle takes into consideration the charged particles that produce an electric field.
- Superposition theorem calculates the force on one charge applied by other charges present in the electric field.
- The electric field that is generated can be calculated through Coulomb’s law.
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Key Terms: Superposition Principle, Electrostatic force, Electric field, Continuous charge distribution, Charge density, Coulomb’s law
What is the superposition principle?
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Superposition Principle is used to calculate the applied total force on a certain point charge by other charges present in the electric field. The created electric field by the charge doesn’t get affected by the presence of any charges. The net calculated electric field is the vector sum of all the charges present there.
- The presence of other charges has no effect on the produced electric field by the point charge charge.
- Net electric field is calculated by Coulomb’s law.
- The superposition concept allows two or more electric fields to be merged.
Read More: Electrostatic Potential
Concept of superposition and calculating force
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To clearly understand the concept of superposition principle three charges q1, q2 and q3 are considered. Here, q2 and q3 charges are applying force on the q1 point charge and now through superposition principle the resultant force on q1 due to the applied force by q2 and q3 is being calculated.
- A vector addition of the forces indicating direction of the force field.
- Resultant force F12 reflects the force applied by q2 on q1 in the presence of charge q3.
\(\vec{F_{12}} = - \frac{1}{4 \pi \epsilon_0} \frac{q_1q_2}{r_{12}} \hat{r_{12}}\)
- Again, similarly F13 will reflect the amount of force applied by q3 on q1 in the presence of q2.
\(\vec{F_{13}} = - \frac{1}{4 \pi \epsilon_0} \frac{q_1q_2}{r_{13}} \hat{r_{13}}\)
- Now, F1, the total force on q1 due to applied forces by q2 and q3 can be calculated as,
\(\vec{F_{1}} = - \frac{1}{4 \pi \epsilon_0} \frac{q_1q_2}{r_{12}} \hat{r_{12}} + \frac{1}{4 \pi \epsilon_0} \frac{q_1q_2}{r_{13}} \hat{r_{13}}\)
\(\vec{F_{1}} = - \frac{4}{4 \pi \epsilon_0} [\frac{q_2}{r_2} \hat{r_{12}} + \frac{q_3}{r_3} \hat{r_{13}}]\)
Continuous charge distribution
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Continuous quantized charges for a variety of reasons are different in working with discrete charges. In continuous charge distribution, all charges are closely packed with each other separated by little distance. This continuous charge distribution can be distributed in three different ways-
- Linear charge distribution
- Surface charge distribution
- Volume charge distribution
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| Related Topics | ||
|---|---|---|
| Electrostatics | Electric Charge | Electrostatic Potential |
| Electric Charges and Fields | Electric Field Formula | Electromagnetic Field |
Linear charge distribution
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The distribution of charges along a small sectional circumference of a straight or circular wire is described as linear charge distribution.
Linear charge distribution
- The symbol λ represents the linear charge distribution.
- A wire's linear charge density is defined as λ = △Q / △I , where Δl is a line element of the wire and ΔQ is the charge present in that line element of the wire.
- Coulombs per meter is the unit of measurement.
Surface charge distribution
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Point charges spread along the surface of a charged conductor surface are described as surface charge distribution.
Surface charge distribution
- Point charge Q is spread on a surface area element S.
- Surface charge density σ by area element can be defined as σ = △Q / △s
- The unit of surface charge density σ is C/m2.
Read More: Charge Density Formula
Volume charge distribution
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Point charge per unit volume of a cube or sphere is referred to as volume charge distribution.
Volume charge distribution
- ρ = △Q / △V, it is defined as the volume charge density or charge density, where ∆Q represents the point charge and ∆V is the element of volume.
- C/m3 is the volume charge density unit.
Coulomb’s Law
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Coulomb's law states that the force of attraction or repulsion between two charged objects is directly proportional to the product of their charges and inversely proportional to the square of their distance. The two charges considered here are point charges.
- Coulomb's law expresses the force between two point charges mathematically.
- Coulomb investigated that the force between two point charges is inversely proportional to the square of their distance, directly proportional to the product of their magnitudes.
- In a vacuum, the amount of force (F) between two point charges q1 and q2 separated by a distance r is given by, F ∝ q1q2 and F ∝ 1/r2 ;Therefore, F ∝ q1q2 / r2 F = k q1q2 / r2. Here, k is proportionality constant and equals 1/4πε0. ε0 is called epsilon not, and it signifies the permittivity of a vacuum.
- The S.I unit value of ε0 is 8.854 × 10-12 C2 N-1 m-2.
Read More: Maxwell’s Equations
Things to Remember
- Total force applied on a given charge is measured by the Superposition Principle.
- Continuous charge distribution can be divided in three categories- Linear charge distribution, Surface charge distribution and Volume charge distribution.
- Linear charge density is defined as λ = △Q / △I, unit is Coulomb/meter.
- Surface charge density σ by area element can be defined as σ = △Q / △s and It is defined as, ρ = △Q / △V and the unit is C/m3.
- Coulomb's law expresses the force between two point charges mathematically.
- The expression for Coulomb’s law F = k q1q2 / r2.
Read More: Electric Charges and Fields Important Pointers PDF
Sample Questions
Ques. What is the dimension of linear charge density? (2 Marks)
Ans. The linear charge density is defined as (Amount of charge / Total length).
As we know, the dimension of electric charge is [I T], while the length dimension is [L]. Then, the linear charge density dimension will be– [I T L-1].
Ques. What are Limitations of Superposition Theorem? (3 Marks)
Ans. The limitations of superposition theorem is as follows -
1) To employ the superposition theorem, all of the components must be linear and as power is not a linear number, this theorem does not apply to it.
2) The total is positive if a contribution from a source has the same direction as the reference direction, so double check is necessary for assigning a reference direction.
3) Non-linear circuits are not covered by the superposition theorem.
4) The superposition theorem necessitates the use of two or more sources in the circuit.
Ques. What is the force between two small-charged spheres with charges of 2 × 10−7 C and 3 × 10−7 C placed 30 cm apart in the air? (5 Marks)
Ans. According to Coulomb’s law, F = (¼ πε0) q1q2/r2, Where q1 and q2 represent the charges and the distance between the charges is denoted by r.
Here, ¼ π ε0 = 9 x 109 N m2 C-2.
permittivity of free space (ε0) = 8.85 x 10-12 (F/m).
The force will be repulsive because both charges are positive.
F12 is the force exerted on charge q1 by charge q2.
Now consider the following: q1 = 2 × 10–7 C,
charge on the first sphere and q2 = 3 × 10–7 C,
Charge on the second sphere.r = 30 cm = 0.3 m, is the distance between the spheres. Now, when we plug the data into the equation we obtain F = q1q2 / 4πε0r2 = (9x 109 x 2 x 10-7 x 3 x 10-7)/(0.3)2 = 6 x 10-3 N.
As a result, the force between the specified charged particles is 6 x 10-3 N.
As the charges have the same nature, i.e. they are both positive. As a result, the force will be repelling.
Ques. A circular annulus of inner radius r and outer radius R has a uniform charge density a. What will be the total charge on the annulus? (3 Marks)
Ans. The annulus has a total surface area of π × (R2 – r2).
Outside radius is R and inside radius is r.
The amount of charge stored on a unit surface area is defined as the surface charge density.
a is the surface charge density.
As a result, the total charge on the annulus is π × a × (R2 – r2).
Ques. The electrostatic force on a small sphere of charge 0.4μC due to another small sphere of charge − 0.8μC in air is 0.2 N. (a) What is the distance between the two spheres? (b) What is the force on the second sphere due to the first? (3 Marks)
Ans. F = 0.2N electrostatic force on the first sphere.
First sphere’s charge q1 = 0.4μC = 0.4 × 10−6C and second sphere’s charge q2 = − 0.8μC = − 0.8 × 10−6C.
The relation describes the electrostatic force between the spheres, F = q1q2/4πε0r2 And, ¼ πε0 = 9 × 109N.m2.C−2 Where, ε0= Permittivity of free space
r2 = q1q2/4πε0F = (04 × 10−6) × (8 × 10−6) × (9 × 10 ) / 0.2 = 144 × 10−4 ,
r = \(\sqrt{144 × 10^{-4}}\) = 0.12m.
The two spheres are separated by 0.12 m.
(b) Equal and Opposite Force operates on the opposite sphere (according to Newton's Third Law). As a result, 0.2 N is attractive.
Ques.What is linear charge distribution? (2 Marks)
Ans. The distribution of charges along a small sectional circumference of a straight or circular wire is described as linear charge distribution. The symbol λ represents the linear charge distribution. A wire's linear charge density is defined as, λ =△Q / △I, where Δl is a line element of the wire and ΔQ is the charge present in that line element.
Ques. What are the various types of charge distribution? Give their formulas. (2 Marks)
Ans. There are three types of charge distribution: Linear charge distribution λ = △Q / △I,
Volume charge distribution ρ = △Q / △V and surface charge distribution σ = △Q / △s.
Ques. A total charge of 10 C is uniformly distributed throughout a solid nonconducting sphere with a radius of 1m. Calculate the sphere's charge density? (2 Marks)
Ans. The sphere's volume is equal to (4/3)πr3., where r is the sphere's radius.
As a result, the charge density is ρ= total charge/[(4/3)πr3.
Using the values as a guide, ρ = 10/[(4/3)πr3] ,ρ= 2.38 C/m3
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