Calculate the electric field of a dipole in an axial position and an endon position.

 Consider an electric dipole consisting of charges –q and +q separated by a small distance 2r in free space. Let P be a point on the axial line of the dipole at a distance x from the centre O of the dipole. (i.e OP = x) Electric field intensity at point P due to +q charge E_1 = 1/(4piepsilon_0).q/(AP^2)  (direction A to P) e_1 = 1/(4piepsilon_0).q/(x+r)^2  (direction A to P)  ...(1) Electric field intensity at point P due to –q charge E_2 = 1/(4piepsilon_0).q/(BP^2)  (direction P to B) E_2 = 1/(4piepsilon_0).q/(x-r)^2  (direction P to B)  ...(2) Since E2 > E1 and they act in opposite directions, resultant field intensity is given by : E = E2 - E1  (direction P to B) = 1/(4piepsilon_0).q/(x-r)^2 - 1/(4piepsilon_0) .q/(x+r)^2 = 1/(4piepsilon_0).q[1/(x-r^2) - 1/(x+r)^2] = 1/(4piepsilon_0).q[((x+r)^2 -(x-r)^2)/((x-r)^2 (x+r)^2)] =1/4piepsilon_0.q[((x+r+x-r)(x+r-x+r))/(xx^2-r^2)^2] = 1/4piepsilon_0.q [(2x.2r)/(x^2-r^2)^2] E= 1/(4piepsilon_0).  (2xp)/(x^2-r^2)^2 [∵ p = 2r.q] If the dipole is short (i.e., r << x) then r2  may be neglected as compared to x2 E = 1/(4piepsilon_0) . (2p)/x^3 The direction of resultant electric field E is along the dipole axis i.e., from –q charge to +q charge. A line passing through the positive and negative charges of the dipole is called the axial line of the electric dipole. Consider a point P on the axial line of a dipole (situated in a vacuum) of length 2a at a distance r from the midpoint O. Electric intensity at P due to –q charge at A, i.e., $$\begin{array}{l}{{E}_{1}}=k\frac{q}{A{{P}^{2}}}=k\frac{q}{{{\left( r+a \right)}^{2}}}\end{array}$$ It is represented in magnitude and direction by $$\begin{array}{l}\overrightarrow{PQ}\end{array}$$, i.e., $$\begin{array}{l}\overrightarrow{{{E}_{1}}.}\end{array}$$ Electric intensity at P due to +q charge at B, i.e., $$\begin{array}{l}{{E}_{2}}=k\frac{q}{B{{P}^{2}}}=k\frac{q}{{{\left( r-a \right)}^{2}}}\end{array}$$ It is represented in magnitude and direction by $$\begin{array}{l}\overrightarrow{PR},\end{array}$$ i.e., $$\begin{array}{l}\overrightarrow{{{E}_{2}}.}\end{array}$$ Let $$\begin{array}{l}\overrightarrow{E}\end{array}$$ be the resultant electric intensity at P. According to the principle of superposition of electric fields, $$\begin{array}{l}\overrightarrow{E}=\overrightarrow{{{E}_{1}}}+\overrightarrow{{{E}_{2}}}\end{array}$$ Since $$\begin{array}{l}\overrightarrow{{{E}_{1}}}\end{array}$$ and E2 act in opposite directions, the resultant electric intensity $$\begin{array}{l}\left( \overrightarrow{E} \right)\end{array}$$ at P due to the dipole is represented by $$\begin{array}{l}\overrightarrow{PS}.\end{array}$$ Clearly, as $$\begin{array}{l}\left| \overrightarrow{{{E}_{2}}} \right|>\left| \overrightarrow{{{E}_{1}}} \right|,\end{array}$$ $$\begin{array}{l}E={{E}_{2}}-{{E}_{1}}=k\left[ \frac{q}{{{\left( r-a \right)}^{2}}}-\frac{q}{{{\left( r+a \right)}^{2}}} \right]\end{array}$$ $$\begin{array}{l}=kq\left[ \frac{{{\left( r+a \right)}^{2}}-{{\left( r-a \right)}^{2}}}{{{\left( r-a \right)}^{2}}{{\left( r+a \right)}^{2}}} \right]=kq\frac{4ra}{{{\left( {{r}^{2}}-{{a}^{2}} \right)}^{2}}}\end{array}$$ Or $$\begin{array}{l}E=k\frac{2\times 2qa\times r}{{{\left( {{r}^{2}}-{{a}^{2}} \right)}^{2}}}=k\frac{2pr}{{{\left( {{r}^{2}}-{{a}^{2}} \right)}^{2}}}\end{array}$$ Since $$\begin{array}{l}\overrightarrow{E}\,and\,\overrightarrow{p}\end{array}$$ are in the same direction, $$\begin{array}{l}\overrightarrow{E}=k\frac{2\overrightarrow{p}\,r}{{{\left( {{r}^{2}}-{{a}^{2}} \right)}^{2}}}=\frac{1}{4\pi {{\in }_{0}}}\frac{2\overrightarrow{p}\,r}{{{\left( {{r}^{2}}-{{a}^{2}} \right)}^{2}}}\end{array}$$ … (1) Special cases: If r is very large as compared to a, i.e., if the dipole is short, a2 can be neglected as compared to r2 and as such, $$\begin{array}{l}\overrightarrow{E}=k\frac{2\,\overrightarrow{p}}{{{r}^{3}}}=\frac{1}{4\pi \,{{\in }_{0}}}\frac{2\overrightarrow{p}}{{{r}^{3}}}\end{array}$$ … (2) E varies as $$\begin{array}{l}1/{{r}^{3}}\end{array}$$ in case of an electric dipole whereas in case of monopole, $$\begin{array}{l}E\propto \frac{1}{r}.\end{array}$$ If the electric dipole is situated in a medium of relative permittivity $$\begin{array}{l}{{\in }_{r}},\end{array}$$ then $$\begin{array}{l}\overrightarrow{E}=k\frac{2\overrightarrow{p}\,r}{{{\in }_{r}}{{\left( {{r}^{2}}-{{a}^{2}} \right)}^{2}}}=\frac{1}{4\pi {{\in }_{0}}{{\in }_{r}}}\frac{2\overrightarrow{p}\,r}{{{\left( {{r}^{2}}-{{a}^{2}} \right)}^{2}}}\end{array}$$ and $$\begin{array}{l}\overrightarrow{E}=k\frac{2\overrightarrow{p}\,r}{{{\in }_{r}}{{r}^{3}}}=\frac{1}{4\pi {{\in }_{0}}{{\in }_{r}}}\frac{2\overrightarrow{p}}{{{r}^{3}}}\end{array}$$ (Short dipole) … (3) Was this answer helpful?       3 (31) Thank you. Your Feedback will Help us Serve you better. Derive an expression for intensity of electric field at a point in broadside position or on an equatorial line of an electric dipole. Electric field intensity at a point on the broad side-on position equatorial line:Consider an electric dipole consisting of two equal but opposite charges -q and +q separated by a vector distance 2l. Let P be a point at a distance r from the centre of the dipole O. The electric intensity at P due to the dipole is the vector sum of the field due to the charge -q at A and +q at B.The resultant intensity is the vector sum of EA and EB.EA and EB can be resolved into two components.The y-components of the field cancel each other because,EA sin  = EB sin , oppositely directed.The x components add up to give the resultant field E.Magnitude of E is,For the right angled triangle OPB,The dipole moment,p = 2lqE = E is directed along PC. The direction of E can be found out by drawing the line of force passing through the point P. The direction of e at P is opposite to the direction of the dipole moment p.That is, parallel to the line joining the two charges and directed from +q to -q.When r>>1, then l2 is very small as compared to r2.Then,