Courses
Courses for Kids
Free study material
Offline Centres
More
Store

# Derive an expression for the intensity of the electric field produced by an electric dipole at a point on its axial line.

Last updated date: 13th Jun 2024
Total views: 401.7k
Views today: 5.01k
Verified
401.7k+ views
Hint: We will use Coulomb’s law equation to derive an expression for the intensity of the electric field. The electric fields will be calculated at a point with different distances. the expression will be the difference of these electric fields being calculated.

Formula used:
$E=\dfrac{1}{4\pi {{\varepsilon }_{0}}}\dfrac{q}{{{r}^{2}}}$

A line passing through the pair of equal and opposite charges of a dipole is called the axial line.
Consider a diagram representing the axial line of a dipole.

Let the charges of a dipole be represented by $-q$and $+q$. Let the distance between the charges be $2a$. Let ‘P’ be a point on the axis line at which the electric field to be determined.
Consider the figure while going through the following steps.
The electric field at a point P with the distance BP is expressed as follows.
Along the line BP due to the charge$+q$.
\begin{align} & {{E}_{B}}=\dfrac{1}{4\pi {{\varepsilon }_{0}}}\dfrac{q}{{{(BP)}^{2}}} \\ & \Rightarrow {{E}_{B}}=\dfrac{1}{4\pi {{\varepsilon }_{0}}}\dfrac{q}{{{(r-a)}^{2}}} \\ \end{align}
The electric field at a point P with the distance PA is expressed as follows.
Along the line PA due to the charge $-q$.
\begin{align} & {{E}_{A}}=\dfrac{1}{4\pi {{\varepsilon }_{0}}}\dfrac{q}{{{(AP)}^{2}}} \\ & \Rightarrow {{E}_{A}}=\dfrac{1}{4\pi {{\varepsilon }_{0}}}\dfrac{q}{{{(r+a)}^{2}}} \\ \end{align}
The net field at the point P is calculated as the difference between the electric fields at the distances BP and PA.
Thus, the net electric field at point P is,
${{E}_{P}}={{E}_{B}}-{{E}_{A}}$
Substitute the expressions of the electric fields at the distances BP and PA in the above equation.
${{E}_{P}}=\dfrac{1}{4\pi {{\varepsilon }_{0}}}\dfrac{q}{{{(r-a)}^{2}}}-\dfrac{1}{4\pi {{\varepsilon }_{0}}}\dfrac{q}{{{(r+a)}^{2}}}$
Take the common terms aside and solve further.
${{E}_{P}}=\dfrac{q}{4\pi {{\varepsilon }_{0}}}\left( \dfrac{1}{{{(r-a)}^{2}}}-\dfrac{1}{{{(r+a)}^{2}}} \right)$
Continue the further calculation.
\begin{align} & {{E}_{P}}=\dfrac{q}{4\pi {{\varepsilon }_{0}}}\left( \dfrac{4ar}{{{({{r}^{2}}-{{a}^{2}})}^{2}}} \right) \\ & \Rightarrow {{E}_{P}}=\dfrac{2qa}{4\pi {{\varepsilon }_{0}}}\left( \dfrac{2a}{{{({{r}^{2}}-{{a}^{2}})}^{2}}} \right) \\ \end{align}
From the diagram it’s clear that, $2aq=p$and the constant value ${}^{1}/{}_{4\pi {{\varepsilon }_{0}}}$can be expressed using a variable ‘k’.
So, we have the expression for the electric field at point P along the line BP as follows.
${{E}_{P}}=\dfrac{2kpr}{{{({{r}^{2}}-{{a}^{2}})}^{2}}}$
Consider a special case, for which the condition is, $2a << r$, the expression for the electric field can be derived as,
${{E}_{P}}=\dfrac{2kp}{{{r}^{3}}}$
Therefore, the expression for the intensity of the electric field produced by an electric dipole at a point on its axial line is ${{E}_{P}}=\dfrac{2kp}{{{r}^{3}}}$.

Note: The formula used to derive the expression is Coulomb’s law for a point charge. This is because we will be solving the electric field at a point, but considering two different distances. If the electric field is given to be within the distance between the charges of a dipole, then, the final expression should be used as discussed above.