Question

(a) An electric dipole of dipole moment $\overrightarrow{p}$ consists of point charges $+q$ and $-q$ separated by a distance $2a$ apart. Deduce the expression for the electric field $\overrightarrow{E}$ due to the dipole at a distance x from the centre of the dipole on its axial line in terms of the dipole moment $\overrightarrow{p}$. Hence show that in the limit $X>>a,\overrightarrow{E}\to 2\overrightarrow{p}/\left(4\pi {ϵ}_0{x}^3\right)$.
(b) Given the electric field in the region $\overrightarrow{E}=2x\hat{i}$, find the net electric flux through the cube and the charge enclosed by it

Answer 1

(a)

$\overrightarrow{p}=2qa\hat{i}$

$\overrightarrow{E_{+q}}=\frac{Kq}{(x-a{)}^2}\hat{i}$

$\overrightarrow{E_{-q}}=-\frac{Kq}{(x+a{)}^2}\hat{i}$

$\overrightarrow{E}=(\frac{Kq}{(x-a{)}^2}-\frac{Kq}{(x+a{)}^2})\hat{i}$

$\overrightarrow{E}=\frac{4Kqax}{(x^2-a^2{)}^2}\hat{i}$

$\overrightarrow{E}=\frac{2Kx\overrightarrow{p}}{(x^2-a^2{)}^2}$

Using binomial expansion,

$\overrightarrow{E}=\frac{2K\overrightarrow{p}}{x^3}(1-\frac{x^2}{a^2}+2\frac{x^4}{a^4}-....)$

Neglecting higher terms,

$\overrightarrow{E}=\frac{2\overrightarrow{p}}{4\pi {\epsilon }_o{x}^3}$

(b)

Since electric field is along x-axis, the only surfaces through which electric flux is non-zero are those parallel to the y-z plane.

At x=a,

${\varphi }_1=\int \overrightarrow{E}.\overrightarrow{dS}$

${\varphi }_1=-(2x{)}_{x=0}\times a^2=0$

${\varphi }_2=\int \overrightarrow{E}.\overrightarrow{dS}$

${\varphi }_2=+(2x{)}_{x=a}\times a^2=2a^3$

Net flux is given by:

$\varphi =2a^3$

By gauss law,

$\varphi =q_{enclosed}/{\epsilon }_o$

$q_{enclosed}=2a^3{\epsilon }_o$