Derive an expression for electric field intensity on perpendicular bisector of dipole and "Also explain its special case."​

Answers (1)

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.

{| \overrightarrow{E_{1}}} |=\frac{1}{4 \pi \epsilon_{0}}* \frac{q}{r^{2}+a^{2}}

{| \overrightarrow{E_{2}}} |=\frac{1}{4 \pi \epsilon_{0}}* \frac{q}{r^{2}+a^{2}}

So \overrightarrow{| {E_{1}}} |=\overrightarrow{| {E_{2}}} |=\overrightarrow{| {E}} |

\begin{aligned} |\vec{E}| &=2 |E_{1}| \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(2a)

\therefore\vec{E}=\frac{-\vec{P}}{4 \pi \epsilon_{0}\left(r^{2}+a^{2}\right)^{3 / 2}}

And 

  •  if r>>a

        then    \vec{E}_{net}=\frac{ -K \vec{P}}{r^{3}}=\frac{ -\vec{P}}{4 \pi \epsilon_{0} r^{3}}   (This is the value of E_{net} when the dipole is placed in the vacuum.)

           

          If the dipole is placed in the medium having the permittivity as \epsilon _m 

           Then \vec{E}_{net}=\frac{- \vec{P}}{4 \pi \epsilon_{m }r^{3}}=\frac{ -\vec{P}}{4 \pi \epsilon _{0} \epsilon _{r} r^{3}} 

Note: Here the direction of the electric field  E is opposite to the direction of  \vec{P}  .

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