Question

The potential due to a short dipole of dipole moment $4.5\times {10}^{-10}Cm$ at a distance $1$ metre from it,

(i) on its axis, (ii) on its perpendicular bisector is :

• A. $0V,5V$
• B. $5V,10V$
• C. $4.05V,0$
• D. $5V,0$

Answer 1

Answer: (C)

$4.05V,0$

The potential of a dipole is given by,
$P=\frac{kpcos\theta }{r^2}$

For a point in the direction of dipole on the axis $\theta =0$
$P_{axis}=kp/r^2=9\times {10}^9\times 4.5\times {10}^{-10}=4.05V$
For a point in the axis perpendicular to dipole $\theta =90$
$P_{perpendicular}=0V$