Question

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

Answer 1

Answer: (D)

Let the length of dipole is AB and mid-point be O.
Electric field at point O by charge $+q$
$E_1=\frac{1}{4\pi {\epsilon }_0}=\frac{q}{{(1/2)}^2}$
$\Rightarrow$ $E_1=\frac{1}{4\pi {\epsilon }_0}\frac{4q}{I^2}$ (along OA)
and electric field at point 0 by charge $(-q)$
$E_2=\frac{1}{4\pi {\epsilon }_0}\frac{q}{{(1/2)}^2}$ $E_2=\frac{1}{4\pi {\epsilon }_0}\frac{4q}{I^2}$ (only OA)
Total electric field at point $O$ $E=E_1+E_2$
$E=2.\left(\frac{1}{4\pi {\epsilon }_0}\frac{4q}{l^2}\right)$
$\Rightarrow$ $E=\frac{1}{4\pi {\epsilon }_0}\frac{8q}{l^2}$ ..(i)
the electric potential of point O by charge +q
$V_1=\frac{1}{4\pi {\epsilon }_0}=\frac{q}{l/2}$
$\Rightarrow$ $V_1=\frac{1}{4\pi {\epsilon }_0}\frac{2q}{l}$
The electric potential at point O by charge $(-q)$
$V_2=\frac{1}{4\pi {\epsilon }_0}.\frac{-q}{l/2}$
$V_2=\frac{1}{4\pi {\epsilon }_0}.\frac{2q}{l}$
$\because$ The total potential at point O $V=V_1+V_2$
$V=0$ ...(ii)
Eq. (i) divide by Eq. (i), we get
$\frac{E}{V}=\infty$