Electric Field Intensity due to a Point Charge
Question: Obtain an expression for electric field intensity due to a point charge (i) when source charge is at the origin, (ii) when source charge is away from origin.
Case 1. When source charge is at the origin.
Consider a static point charge Q at the origin O (0, 0, 0). Let $q_0$ be the test charge placed in free space at a point P, distant r from O in the electric field of the charge Q.
According to Coulomb's law, force experienced by $q_0$ due to charge Q is given by
$\vec{F} = \frac{1}{4\pi \epsilon_0} \frac{Qq_0}{r^2} \hat{r}$...(i)
where $\hat{r}$ is a unit vector directed along OP.
Now, electric field intensity at point P due to point charge Q is given by
$\vec{E} = \frac{\vec{F}}{q_0} = \frac{1}{q_0} \times \frac{1}{4\pi \epsilon_0} \frac{Qq_0}{r^2} \hat{r}$
$\vec{E}= \frac{1}{4\pi \epsilon_0} \frac{Q}{r^2} \hat{r}...(ii)$
Magnitude of the electric field intensity is given by
$E = |\vec{E}| = \frac{1}{4\pi \epsilon_0} \frac{Q}{r^2}$...(iii)
Thus, the magnitude of the electric field intensity at a point due to a point source charge is (i) directly proportional to the magnitude of the charge and (ii) inversely proportional to the square of the distance of the point from the source charge.
The direction of electric field intensity $\vec{E}$ is along OP i.e., along the increasing direction of $\vec{r}$ or radially outwards.
When $r \rightarrow \infty$, then $E \rightarrow 0$. Thus, electric field due to a point source charge has infinite range.
The graph between electric field intensity (E) due to a point source charge and the square of the distance ($r^2$) of test charge from the source charge is
The graph between E (electric field intensity due to a point source charge) and $1/r^2$ (where r is distance of test charge from the source charge).
Case 2. When source charge is away from the origin.
Consider a static point charge Q placed at A and a test charge $q_0$ be placed in the electric field of charge Q at an observation point P such that
$\vec{OA} = \vec{r}_1$ and $\vec{OP} = \vec{r}_2$
Coulomb's force experienced by the charge $q_0$ due to charge Q is given by
$\vec{F} = \frac{1}{4\pi \epsilon_0} \frac{Qq_0}{|\vec{AP}|^2} \hat{r}_{AP}...(i)$
where $\hat{r}_{AP}$ is the unit vector along AP
$\vec{AP} = (\vec{r}_2 - \vec{r}_1)$ or $|\vec{AP}| = |\vec{r}_2 - \vec{r}_1 ...(ii)$
Also $\vec{AP} = |\vec{AP}| \hat{r}_{AP}$
so $\hat{r}_{AP} = \frac{\vec{AP}}{|\vec{AP}|} = \frac{(\vec{r}_2 - \vec{r}_1)}{|\vec{r}_2 - \vec{r}_1|}$ ...(iii)
Substituting the value of eqns. (ii) and (iii) in eqn. (i), we get
$\vec{F} = \frac{1}{4\pi \epsilon_0} \frac{Qq_0}{|\vec{r}_2 - \vec{r}_1|^2} \times \frac{(\vec{r}_2 - \vec{r}_1)}{|\vec{r}_2 - \vec{r}_1|}$
$\vec{F} = \frac{1}{4\pi \epsilon_0} \frac{Qq_0(\vec{r}_2 - \vec{r}_1)}{|\vec{r}_2 - \vec{r}_1|^3}$ ...(iv)
Now, electric field due to the charge Q at point P is given by
$\vec{E} = \frac{\vec{F}}{q_0} = \frac{1}{4\pi \epsilon_0} \frac{Q(\vec{r}_2 - \vec{r}_1)}{|\vec{r}_2 - \vec{r}_1|^3}$
The direction of electric field $\vec{E}$ is along AP i.e. away from a positive source charge Q.