Question : What is meant by 'continuous charge distribution'? Define: linear charge density, surface charge density and volume charge density. Write their units in SI. Find Force and Field due to continuous charge distribution - Param Himalaya
Continuous charge distribution
A system of closely spaced electric charges forms a continuous charge distribution. At macroscopic scale, the number of charges is very large. Practically at this level, it is not feasible to count the number of charges, so quantization of charges can be easily ignored. In other words, it can be said that at macroscopic level, the charges are supposed to be continuously distributed.
(i) Linear charge density (λ): Linear charge density is defined as charge per unit length.
If a charge q is uniformly distributed along a line or a wire of length l, then the linear charge density of the line or the wire is given by:
$\lambda = \frac{q}{l}$
The SI unit of $\lambda$ is coulomb/metre (C m$^{-1}$).
The charge on a small element of length $dl$ of an infinite charged line having linear charge density of the surface $\lambda$ is given by
$dq = \lambda dl$
(ii) Surface charge density ($\sigma$):
Surface charge density is defined as charge per unit area.
If a charge $q$ is uniformly distributed on a surface of area $S$, then the surface charge density of the surface $\sigma$ is given by
$\sigma = \frac{q}{S}$
The SI unit of $\sigma$ is coulomb/metre$^2$ (C m$^{-2}$).
The charge on a small element of area $dS$ of a surface having surface charge density $\sigma$ is given by
$dq = \sigma dS$
(iii) Volume charge density ($\rho$): Volume charge density is defined as charge per unit volume.
If a charge $q$ is uniformly distributed throughout a volume $V$, then the volume charge density is given by
$\rho = \frac{q}{V}$
The S.I. unit of $\rho$ is coulomb/metre$^3 \ Cm^{-3}$
The charge on a small element of volume $dV$ of a surface of volume $V$ having volume charge density $\rho$ is given by
$dq = \rho dV$
1. Electric field intensity due to linear charge distribution
When the charge is uniformly distributed over a line (straight or curved), we call it linear charge distribution. For example, charge on a long wire or a ring etc.
Consider a straight wire AB of length $l$ on which charge $q$ is distributed uniformly. Let $\lambda$ be the linear charge density (i.e., charge/length) of the wire.
Consider a small element of length $dl$ on the wire. Charge on the element of length $dl$ is given by,
$dq = \lambda dl \qquad \cdots (i)$
Electrostatic force on a test charge $q_0$ placed at a point P at a distance $r$ from the small element of length is given by,
$d\vec{F} = \frac{1}{4\pi\epsilon_0} \frac{dq \times q_0}{r^2} \hat{r}$
Therefore, total force on the test charge $q_0$ due to charge on the whole wire is given by
$\vec{F} = \int d\vec{F} = \int \frac{1}{4\pi\epsilon_0} \frac{dq \times q_0}{r^2} \hat{r}$
Using equation (i), we have;
$\vec{F} = \int_l \frac{1}{4\pi\epsilon_0} \frac{\lambda dl \times q_0}{r^2} \hat{r}$
$\vec{F} = \frac{q_0}{4\pi\epsilon_0} \int_l \frac{\lambda}{r^2} dl \hat{r} \qquad \cdots (ii)$
$\vec{F} = \frac{q_0}{4\pi\epsilon_0} \int_l \frac{\lambda dl}{r^2} \hat{r}$
Electric field intensity $\vec{E}$ is given by
$\vec{E} = \frac{\vec{F}}{q_0} = \frac{1}{4\pi\epsilon_0} \int_l \frac{\lambda dl}{r^2} \hat{r}$
which is the electric field intensity due to a linear charge distribution at a distance $r$.
2. Electric field intensity due to surface charge distribution :
When the charge is continuously distributed uniformly over a surface, we call it surface charge distribution. For example, charge on a membrane or a film or a plane sheet.
Consider a plane sheet of surface area S on which charge q is distributed uniformly
Let $\sigma$ = surface charge density (charge/area) of the surface.
Now consider a small element of area $dS$, then charge on the small element is given by,
$dq = \sigma dS \qquad \cdots (i)$
Electrostatic force on charge $q_0$ placed at a point P at a distance $r$ from the element is given by
$d\vec{F} = \frac{1}{4\pi\epsilon_0} \frac{dq \times q_0}{r^2} \hat{r}$
$d\vec{F} = \frac{1}{4\pi\epsilon_0} \frac{\sigma dS q_0}{r^2} \hat{r}$
$\therefore$ Total force acting on $q_0$ due to the charge on the whole surface is given by
$\vec{F} = \int_S d\vec{F}$
$\vec{F}= \int_S \frac{1}{4\pi\epsilon_0} \frac{\sigma dS q_0}{r^2} \hat{r}$
$\vec{F} = \frac{q_0}{4\pi\epsilon_0} \int_S \frac{\sigma}{r^2} dS \hat{r} \qquad \cdots (ii)$
Force experienced by a test charge $q_0$ due to surface charge distribution is given by
$\vec{F} = \frac{q_0}{4\pi\epsilon_0} \int_S \frac{\sigma dS}{r^2} \hat{r}$
Therefore, electric field intensity $\vec{E}$ is given by
$\vec{E} = \frac{\vec{F}}{q_0} = \frac{1}{4\pi\epsilon_0} \int_S \frac{\sigma dS}{r^2} \hat{r}$
which is the electric field intensity due to a surface charge distribution at a distance $r$.
3. Electric field intensity due to volume charge distribution
When the charge is continuously distributed over a volume, then we call it volume distribution of charge. For example, charge on a lump, brick etc.
Consider a surface of volume $V$ having charge $q$ which is uniformly distributed over it. Let $\rho$ be the volume charge density (i.e., charge/volume) of the surface.
Now consider a small element of volume $dV$, then the charge on the element is given by
$dq = \rho dV \qquad \cdots (i)$
Now, electrostatic force acting on $q_0$ due to the charge on the element is given by
$d\vec{F} = \frac{1}{4\pi\epsilon_0} \frac{dq \times q_0}{r^2} \hat{r}$
$d\vec{F}= \frac{1}{4\pi\epsilon_0} \frac{\rho dV q_0}{r^2} \hat{r}$
Therefore, total force acting on the charge $q_0$ due to the charge on the whole volume is given by
$\vec{F} = \int_V d\vec{F} = \int_V \frac{1}{4\pi\epsilon_0} \frac{\rho dV q_0}{r^2} \hat{r}$
or
$\vec{F} = \frac{q_0}{4\pi\epsilon_0} \int_V \frac{\rho}{r^2} dV \hat{r}$
Force experienced by a test charge $q_0$ due to volume charge distribution is given by,
$\vec{F} = \frac{q_0}{4\pi\epsilon_0} \int_V \frac{\rho dV}{r^2} \hat{r}$
Therefore, electric field intensity $\vec{E}$ is given by,
$\vec{E} = \frac{\vec{F}}{q_0} = \frac{1}{4\pi\epsilon_0} \int_V \frac{\rho dV}{r^2} \hat{r}$
which is the electric field intensity due to a volume charge distribution at a distance $r$.