Electric Field Of A Dipole - Practice Questions & MCQ

Electric Field Intensity due to an Electric Dipole at a Point on the Axial Line

As shown in the above figure We want to find out Electric Field Intensity due to an Electric Dipole at a Point M which is on axial line and at a distance r from the centre of a dipole.

Where $E_1$ and $E_2$ is the Electric Field Intensity at M due to $-q$ and $+q$ charges respectively.
The intensities $E_1$ and $E_2$ are along the same line but in opposite directions.

$$
\begin{aligned}
& E_1=\frac{k q}{(r+a)^2} \\
& E_2=\frac{k q}{(r-a)^2} \\
& E_{n e t}=E_2-E_1
\end{aligned}
$$

$$
E_{n e t}=\frac{K q}{(r-a)^2}-\frac{K q}{(r+a)^2}=\left[\frac{4 \text { Kqar }}{\left(r^2-a^2\right)^2}\right]
$$

Using $P=q(2 a)$

$$
E_{n e t}=\left[\frac{2 K P r}{\left(r^2-a^2\right)^2}\right]
$$

- For short/ldeal dipole (i.e r>>a)
then

$$
\vec{E}_{n e t}=\frac{2 K \vec{P}}{r^3}=\frac{2 \vec{P}}{4 \pi \epsilon_0 r^3}
$$

(This is the value of $E_{\text {net }}$ when the dipole is placed in the vacuum.)

If the dipole is placed in the medium having the permittivity as $\epsilon_m$

$$
\text { Then } \vec{E}_{n e t}=\frac{2 \vec{P}}{4 \pi \epsilon_m r^3}=\frac{2 \vec{P}}{4 \pi \epsilon_0 \epsilon_r r^3}
$$

Note: The direction of the electric field E is in the direction of $\vec{P}$.
i.e Angle between $\mathrm{E}_{\mathrm{axi}}$ and $\vec{P}$ is $0^0$.

Electric Field Intensity due to an Electric Dipole at a Point on the Equitorial line.

As shown in the above figure We want to find out Electric Field Intensity due to an Electric Dipole at a Point M which is on the Equitorial line and at a distance r from the center of a dipole.

Where $E_1$ and $E_2$ is the Electric Field Intensity at M due to $-q$ and $+q$ charges respectively.

$$
\begin{aligned}
&\left|\overrightarrow{E_1}\right|=\frac{1}{4 \pi \epsilon_0} * \frac{q}{r^2+a^2} \\
&\left|\overrightarrow{E_2}\right|=\frac{1}{4 \pi \epsilon_0} * \frac{q}{r^2+a^2} \\
& \text { So }\left|\overrightarrow{E_1}\right|=\left|\overrightarrow{E_2}\right|=|\vec{E}| \\
&|\vec{E}|=2\left|E_1\right| \cos \theta \\
&=\frac{2}{4 \pi \epsilon_0} \cdot \frac{q}{\left(r^2+a^2\right)} \cos \theta \\
&=\frac{2}{4 \pi \epsilon_0} \cdot \frac{q}{\left(r^2+a^2\right)} \frac{a}{\sqrt{r^2+a^2}} \\
&=\frac{q \times 2 a}{4 \pi \epsilon_0\left(r^2+a^2\right)^{3 / 2}}
\end{aligned}
$$

Using $P=q(2 a)$

$$
\therefore \vec{E}=\frac{-\vec{P}}{4 \pi \epsilon_0\left(r^2+a^2\right)^{3 / 2}}
$$

And
- For short/ldeal dipole (i.e $\mathrm{r} \gg \mathrm{a}$ )
then

$$
\vec{E}_{n e t}=\frac{-K \vec{P}}{r^3}=\frac{-\vec{P}}{4 \pi \epsilon_0 r^3}
$$

(This is the value of $E_{\text {net }}$ when the dipole is placed in the vacuum.)

If the dipole is placed in the medium having the permittivity as $\epsilon_m$

$$
\vec{E}_{\text {Thet }}=\frac{-\vec{P}}{4 \pi \epsilon_m r^3}=\frac{-\vec{P}}{4 \pi \epsilon_0 \epsilon_r r^3}
$$
 

$
\text { i.e Angle between Eequi and } \vec{P} \text { is } 180^{\circ} \text {. }
$