Let us take a very light rod AB (as shown in the figure 1) of negligible mass and hang three mass ‘m’ each. So the total mass of the rod will be ‘3m’. But the mass of each part of the rod is not uniform, some parts will have more mass like CD and some parts like EF will have less mass.
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Figure-1
This is because masses are concentrated only in some parts and this case is known as discrete mass distribution. Similarly, if we take a rod PQ (as shown in figure 2) which has the mass equally distributed throughout the rod PQ. Now whether we take the part RS or TU the amount of mass remains the same for the same length and this case is called continuous mass distribution.
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Figure-2
Similarly, if we keep charged particles in the place of masses, then again there will be two types of distribution -
Discrete Charge Distribution
Continuous Charge Distribution
In case of discrete charge distribution, the charges are at distance from each other. So all the properties of the charges can be individually recognised. For example, if there are charges q1, q2, q3……. up to qn then the total charges can be calculated by using simple algebra. Also, we can find the electric field due to all the charges individually and total electric field can be calculated with the help of the principle of superposition.
But in case of Continuous charge distribution, the charges are arranged in such a way that the charges are uniformly distributed along the conductor and all the charges are closely packed.
There are three types of Continuous charge distribution -
Linear Charge Distribution
Surface Charge Distribution
Volume Charge Distribution
Let us discuss all the types of Continuous charge distribution one by one.
If the charge is distributed uniformly along the length of the conductor i.e., along the line then it is called Linear Charge Distribution. For example - the distribution of charges along a straight thin wire or the distribution of charges along the circumference of a circle shape wire. It is represented by linear charge density which has the symbol - ‘$\mathrm{\lambda}$’ which is pronounced as Lambda and can be defined as charge per unit length. It can be written as -
$$\mathrm{\Rightarrow\:\lambda\:=\:\frac{dq}{dl}}$$
Where, dq is the amount of charge distributed along the small length dl. So, the charge over the length ‘dl’ will be -
$$\mathrm{dq \:=\: \lambda.dl}$$
So if we want to calculate the total charge ‘q’ along the line then it will be the integration of the product of linear charge density and indefinite small length, which can be written as -
$$\mathrm{\Rightarrow\:q\:=\:\int\:\:\:dq\:=\:\int\:\:\:\lambda. dl}$$
The unit of the linear charge density will be - $\mathrm{Cm^{-1}}$. The figure 3 will show the two case of linear charge distribution, first is the linear charge distribution on straight wire and the second is the linear charge distribution on circumference of the circular wire –
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Figure-3
If the charges are uniformly distributed over the surface of anybody then it is called Surface Charge Distribution. Here the surface will indicate the area covered. For example, the charge is distributed uniformly over a thin sheet. It is represented by linear charge density which has the symbol ‘$\mathrm{\sigma}$’ which is pronounced as Sigma. So the Surface charge density can be written as total charge per unit area. It can be written as -
$$\mathrm{\Rightarrow\:\sigma\:=\:\frac{dq}{dA}}$$
Where the dq is the charge uniformly distributed over the small area dA. So the dq charge can be written as -
$$\mathrm{\Rightarrow\:dq\:=\:\sigma.dA}$$
So if we want to calculate the total charge uniformly distributed over the surface then it will be the integration of product surface charge density and the small area. It can be written as -
$$\mathrm{\Rightarrow\:q\:=\:\int\:dq\:=\:\int\:\sigma.dA}$$
The unit of $\mathrm{\sigma}$ is $\mathrm{C.m^{-2}}$
The figure 4 will show the example of surface charge distribution -
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Figure-4
If the charges are uniformly distributed over the volume of any body then it is called Volume Charge Distribution. For example, the charge is distributed uniformly in a solid cylinder or solid sphere. It is represented by linear charge density which has the symbol ‘$\mathrm{\rho}$’ which is pronounced as Rho. So the Volume charge density can be written as total charge per unit volume. It can be written as -
$$\mathrm{\Rightarrow\:\rho\:=\:\frac{dq}{dV}}$$
Where the dq is the charge uniformly distributed in a small volume dV. So the dq charge can be written as -
$$\mathrm{dq\:=\:\rho.dV}$$
So if we want to calculate the total charge uniformly distributed in the volume then it will be the integration of product volume charge density and the small volume. It can be written as -
$$\mathrm{\Rightarrow\:q\:=\:\int\:dq\:=\:\int\:\rho.dV}$$
The unit of $\mathrm{\rho}$ is $\mathrm{C.m^{-3}}$. The figure 5 will show the example of volume charge distribution -
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Figure-5
Let us understand how we are going to calculate the force exerted by the charge distribution on any other charge. Here we are taking a point charge for the calculation of force for our reference. Let us start with the linear charge distribution -
Let us assume a thin wire having a linear charge density ‘$\mathrm{\lambda}$’. Let us take a small length of wire ‘dl’ and have a charge ‘dq’. Let us take a point charge qo at a distance ‘r’. Then according to the Coulomb's law -
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Figure-6
$$\mathrm{d\overrightarrow{F}\:=\:\frac{q_0.dq}{4\pi \epsilon_0}.\frac{1}{r^2}.\hat{r}}$$
If we are writing the $\mathrm{dq = \lambda.dl}$, then we can write the above equation -
$$\mathrm{\Rightarrow\:d\overrightarrow{F}\:=\:\frac{q_0.\lambda.dl}{4\pi \epsilon_0}.\frac{1}{r^2}.\hat{r}}$$
Let us assume a thin sheet having a surface charge density ‘$\mathrm{\sigma}$’. Let us take a small area ‘dS’ and have a charge ‘dq’. Let us take a point charge qo at a distance ‘r’. Then according to the Coulomb's law -
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Figure-7
$$\mathrm{d\overrightarrow{F}\:=\:\frac{q_0.dq}{4\pi \epsilon_0}.\frac{1}{r^2}.\hat{r}}$$
If we are writing the $\mathrm{dq = \sigma.dA}$, then we can write the above equation -
$$\mathrm{\Rightarrow\:d\overrightarrow{F}\:=\:\frac{q_0.\sigma.ds}{4\pi \epsilon_0}.\frac{1}{r^2}.\hat{r}}$$
Let us assume a body having volume ‘V’ and volume charge density ‘$\mathrm{\rho}$’. Let us take a small area ‘dV’ and have a charge ‘dq’. Let us take a point charge qo at a distance ‘r’. Then according to the Coulomb's law -
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Figure-8
$$\mathrm{d\overrightarrow{F}\:=\:\frac{q_0.dq}{4\pi \epsilon_0}.\frac{1}{r^2}.\hat{r}}$$
If we are writing the $\mathrm{dq = \rho.dV}$, then we can write the above equation -
$$\mathrm{\Rightarrow\:d\overrightarrow{F}\:=\:\frac{q_0.\rho.dV}{4\pi \epsilon_0}.\frac{1}{r^2}.\hat{r}}$$
Q1. What do you mean by continuous charge distribution?
Ans - Continuous charge distribution means that the charges are distributed uniformly on the body. The amount of charge per unit volume or area or length will remain the same in this case.
Q2. What do you mean by the Linear charge distribution?
Ans - If the charges are distributed uniformly in a line or over the length of the body then it is called Linear charge distribution.
Q3. What is Surface charge distribution?
Ans - If the charges are distributed over the surface or area of the body then it is called as surface charge distribution.
Q4. What is volume charge distribution?
Ans - If the charges are distributed in the volume of the body then it is called Volume charge distribution.
Q5. What is the equation of differential force exerted by a wire having linear charge density ‘$\mathrm{\lambda}$’ by the small length ‘dl’ on a point charge (qo) at a distance ‘r’ in the vacuum?
Ans - The amount of force exerted by small length (dl) of a wire having linear charge density ‘$\mathrm{\lambda}$’ in vacuum on $\mathrm{q_o}$ will be -
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$$\mathrm{d\overrightarrow{F}\:=\:\frac{q_0.\lambda.dl}{4\pi \epsilon_0}.\frac{1}{r^2}.\hat{r}}$$
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