question_answer

Find the ratio of electric field to potential EIV at the midpoint of electric dipole, if the distance between the charges is \[l\].

A)  \[1/l\]               

B)  \[l\]

C)  \[2/l\]               

D)  None of these

Correct Answer: D

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 \[\frac{E}{V}=\infty \]