question_answer

Deduce an expression for the electric potential due to an electric dipole at any point on its axis. Mention one contrasting feature of electric potential of a dipole at a point as compared to that of due to a single charge.

Answer:

Electrostatic potential due to a dipole at same angle from its centre. Let us consider an electric dipole consisting of charges \[+\,q\] and \[-\,q\] separated by a distance 2a. The dipole moment \[|p|=q\times 2a.\] Let O be the centre of the dipole, P be any point near the electric dipole inclined at an angle q as shown in the figure. Let P be the point at which electric potential is required. Potential at \[P\] due to \[+\,q\]charge, \[{{V}_{1}}=\frac{q}{4\pi {{\varepsilon }_{0}}{{r}_{1}}}.\] Potential at due to −q charge, \[{{V}_{2}}=\frac{-\,q}{4\pi {{\varepsilon }_{0}}{{r}_{2}}}.\] As potential is related to work done by the field, electrostatic potential also follows the superposition principle. Therefore, potential at P due to the dipole, \[{{V}_{p}}={{V}_{1}}+{{V}_{2}}=\frac{q}{4\pi {{\varepsilon }_{0}}}\,\,\left[ \frac{1}{{{r}_{1}}}-\frac{1}{{{r}_{2}}} \right]\]       ?.. (i) Now, by geometry,             \[r_{1}^{2}={{r}^{2}}+{{a}^{2}}-2ar\,\cos \,\theta \] Similarly, \[r_{2}^{2}={{r}^{2}}+{{a}^{2}}-2ar\,\cos \,(180{}^\circ -\theta )\] Or         \[r_{2}^{2}={{r}^{2}}+{{a}^{2}}+2ar\,\cos \,\theta \] \[[\therefore \,\,cos\,(180{}^\circ -\theta )=-cos\,\theta ]\] And      \[r_{1}^{2}={{r}^{2}}\,\left( 1+\frac{{{a}^{2}}}{{{r}^{2}}}+\frac{2a}{r}\,\,\cos \,\theta  \right)\] If \[r>>a,\,\,\frac{a}{r}\] is small. Therefore, \[\frac{{{a}^{2}}}{{{r}^{2}}}\] can be neglected. \[r_{1}^{2}={{r}^{2}}\,\left( 1-\frac{2a}{r}\,\,\cos \,\theta  \right)\] \[\Rightarrow \] \[{{r}_{1}}=r\,\,{{\left( 1-\frac{2a}{r}\,\,\cos \,\theta  \right)}^{1/2}}\] Or         \[\frac{1}{{{r}_{1}}}=\frac{1}{r}\,\,{{\left( 1-\frac{2a}{r}\,\,\cos \,\theta  \right)}^{-1/2}}\] Similarly, \[\frac{1}{{{r}_{2}}}=\frac{1}{r}\,\,{{\left( 1+\frac{2a}{r}\,\,\cos \,\theta  \right)}^{-1/2}}\] Putting these values in Eq. (i), we obtain             \[{{V}_{P}}=\frac{q}{4\pi {{\varepsilon }_{0}}}\,\,\left[ \frac{1}{r}\,\,{{\left( 1-\frac{2a}{r}\,\,\cos \,\theta  \right)}^{-\frac{1}{2}}}-\frac{1}{r}\,\,{{\left( 1+\frac{2a}{r}\,\,\cos \,\theta  \right)}^{-\frac{1}{2}}} \right]\] Using Binomial theorem,\[[{{(1+x)}^{n}}=1+nx,\,\,x<<1]\]and retaining terms up to the first order in \[\frac{a}{r},\]we obtain \[{{V}_{p}}=\frac{q}{4\pi {{\varepsilon }_{0}}r}\,\,\left[ \left( 1-\frac{a}{r}\,\,\cos \,\theta  \right)-\left( 1-\frac{a}{r}\,\,\cos \,\theta  \right) \right]\]      \[=\frac{q}{4\pi {{\varepsilon }_{0}}r}\,\,\left[ 1-\frac{a}{r}\,\,\cos \,\theta -1+\frac{a}{r}\,\,\cos \,\theta  \right]\] \[{{V}_{p}}=\frac{q}{4\pi {{\varepsilon }_{0}}r}\,\left( \frac{2a\cos \,\theta }{r} \right)=\frac{q\times 2a\,\cos \,\theta }{4\pi {{\varepsilon }_{0}}{{r}^{2}}}\] \[{{V}_{p}}=\frac{p\,\cos \,\theta }{4\pi {{\varepsilon }_{0}}{{r}^{2}}}\]                      \[[\therefore \,\,p=q\times 2a]\] As         \[p\,\cos \,\theta =p.\hat{r}\] Where, \[\hat{r}\] is a unit vector along OP = r. \[\therefore \] Electrostatic potential at point P due to a short dipole \[(a<<r)\] is given by \[V=\frac{p.\hat{r}}{4\pi {{\varepsilon }_{0}}{{r}^{2}}}\] The potential depends just not on the position vector r, but also on the angle between the position vector r and the dipole moment p. The electric potential due to an electric dipole at point P varies inversely with square of r, i.e. the distance of point P from the centre of the dipole. The potential on the dipole axis, \[\theta =0{}^\circ \,\,or\,\,\pi \]             \[\therefore \]      \[\pm \,\frac{p}{4\pi {{\varepsilon }_{0}}{{r}^{2}}}\]             Positive sign for \[\theta =0{}^\circ \] and negative sign for \[\theta =\pi .\] In the equatorial plane, \[\theta =\pi /2\] \[cos\,\theta =cos\,(\pi /2)=0\]      \[\text{ }(\therefore \,\,V=0)\] Thus, electrostatic potential at any point in the equatorial plane of dipole is zero. Differences between electric potential due to an electric dipole and due to a single charge are given as below: (i) The potential due to a dipole depends not just on \[r\] but also on the angle between the position vector r and dipole moment vector p. (ii) The electric potential due to dipole falls off at a distance of \[r\] as \[1/{{r}^{2}}\] not as \[1/r,\] which is a characteristic of the potential due to single charge.