Question

Find the electric potential and then electric field due to an electric dipole by differential relationship between field and potential

Answer 1

let's find potential of electric dipole at an arbitrary point P. at P, the positive point charge sets up potential $V_{+}$ and negative point charge sets up potential $V_{-}$.

so the net potential at P is given by

V = $V_{+}+V_{-}$

= $\frac{q}{4\pi\epsilon_0}\left(\frac{1}{r_{(+)}}-\frac{1}{r_{(-)}}\right)$

= $\frac{q}{4\pi\epsilon_0}\frac{r_{(-)}-r_{(+)}}{r_{(+)}r_{(-)}}$….......(1)

here $r_{(-)}-r_{(+)}\approx dcos\theta$

and $r_{(+)}r_{(-)}\approx r^2$

if we substitute these quantities into equation (1)

we get,

$V=\frac{q}{4\pi\epsilon_0}\frac{dcos\theta}{r^2}$

similarly, let's find electric field due to electric dipole at a point P, a distance z from the midpoint of the dipole and on its central axis, which is called the dipole axis.

electric field due to positive charge , $E_{(+)}=\frac{q}{4\pi\epsilon_0r_{(+)}^2}$

and electric field due to negative charge, $E_{(-)}=\frac{q}{4\pi\epsilon_0r_{(-)}^2}$

so, net electric field , E = $E_{(+)}-E_{(-)}$

= $\frac{q}{4\pi\epsilon_0}\left[\frac{1}{(z-d/2)^2}-\frac{1}{(z+d/2)^2}\right]$

= $\frac{q}{4\pi\epsilon_0}\frac{2dz}{(z^2-d^2/4)^2}$

as z >> d , so z² ≈ (z² - d²/4)

so, $E=\frac{q2d}{4\pi\epsilon_0z^3}$

we know, q.d = P

so, $E=\frac{2P}{4\pi\epsilon_0z^3}=\frac{2\kappa P}{z^3}$

relation between electric field and electric potential is ...

$E_s=-\frac{\delta V}{\delta s}$

Answer 2

Answer:

Explanation:

if cosΦ = 1 , resultant will be maximum i.e., |A| + |B|

and if cosΦ = -1 , resultant will be minimum i.e., |A| - |B|

so, |A| - |B| ≤ (A + B) ≤ |A| + |B|

similarly resultant of f1 and f2 is given by,

f2 - f1 ≤ R ≤ f1 + f2 [ as f1 < f2 ]

on comparing we get,

f2 - f1 = 8 and f1 + f2 = 12

after solving we get, f1 = 2 and f2 = 10