Q. Find the ratio of electric field to potential EIV at the midpoint of electric dipole, if the distance between the charges is $ l $ .
Rajasthan PETRajasthan PET 2008
Solution:
Let the length of dipole is AB and mid-point be O.
Electric field at point O by charge $ +q $
$ {{E}_{1}}=\frac{1}{4\pi {{\varepsilon }_{0}}}=\frac{q}{{{(1/2)}^{2}}} $
$ \Rightarrow $ $ {{E}_{1}}=\frac{1}{4\pi {{\varepsilon }_{0}}}\frac{4q}{{{I}^{2}}} $ (along OA)
and electric field at point 0 by charge $ (-q) $
$ {{E}_{2}}=\frac{1}{4\pi {{\varepsilon }_{0}}}\frac{q}{{{(1/2)}^{2}}} $ $ {{E}_{2}}=\frac{1}{4\pi {{\varepsilon }_{0}}}\frac{4q}{{{I}^{2}}} $ (only OA)
Total electric field at point $ O $ $ E={{E}_{1}}+{{E}_{2}} $
$ E=2.\left( \frac{1}{4\pi {{\varepsilon }_{0}}}\frac{4q}{{{l}^{2}}} \right) $
$ \Rightarrow $ $ E=\frac{1}{4\pi {{\varepsilon }_{0}}}\frac{8q}{{{l}^{2}}} $ ..(i)
the electric potential of point O
by charge +q
$ {{V}_{1}}=\frac{1}{4\pi {{\varepsilon }_{0}}}=\frac{q}{l/2} $
$ \Rightarrow $ $ {{V}_{1}}=\frac{1}{4\pi {{\varepsilon }_{0}}}\frac{2q}{l} $
The electric potential at point O by charge $ (-q) $
$ {{V}_{2}}=\frac{1}{4\pi {{\varepsilon }_{0}}}.\frac{-q}{l/2} $
$ {{V}_{2}}=\frac{1}{4\pi {{\varepsilon }_{0}}}.\frac{2q}{l} $
$ \because $ The total potential at point O $ V={{V}_{1}}+{{V}_{2}} $
$ V=0 $ ...(ii)
Eq. (i) divide by Eq. (i), we get
$ \frac{E}{V}=\infty $
