Question

(a) Derive a relation for electric field due to an electric dipole at a point on the equatorial plane of the electric dipole. Draw necessary diagram.
(b) An electric dipole of charge $\pm 1$ $\mu C$ exists inside a spherical Gaussian surface of radius $1cm$. Write the value of outgoing flux from the Gaussian surface.
(c) Potential on the surface of a charged spherical shell of radius $10cm$ is $10V$. Write the value of potential at $5cm$ from its centre.

Answer 1

(a) : Electric field at point P due to each charge of dipole is shown in the figure.

$E=\frac{kq}{x^2}$

From figure, $x=\sqrt{d^2+r^2}$

$⟹$ $E=\frac{kq}{d^2+r^2}$

Electric field component at P in y direction $E_y=E\sin \theta -E\sin \theta =0$

Electric field component at P in x direction $E_x=-2E\cos \theta$ where $\theta =\frac{d}{x}=\frac{d}{\sqrt{d^2+r^2}}$

$⟹$ $E_x=\frac{-2kqd}{(d^2+r^2{)}^{3/2}}$

Dipole moment of dipole $P=q\left(2d\right)$

$⟹$ Electric field at P $E_P=E_x=\frac{-kP}{(d^2+r^2{)}^{3/2}}$

Negative sign shows that electric field at a point in equatorial plane points in a direction opposite to that of electric dipole moment.

(b) : Total charge enclosed by the spherical Gaussian surface $q_{enc}=1\mu C-1\mu C=0$

Outgoing flux $\varphi =\frac{q_{enc}}{{ϵ}_o}=0$

(c) : The magnitude of potential at a point inside the charged spherical shell is equal to that at its surface because electric field inside that spherical shell is zero.

Potential at the surface $V_s=10$ V (Given)

Thus potential at a point $5cm$ from its centre $V^{\prime }=V_s=10$ V