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Two charges $2 \mu \mathrm{C}$ and $-2 \mu \mathrm{C}$ are placed at points $\mathrm{A}$ and $B 6 \mathrm{~cm}$ apart.
(a) Identify an equipotential surface of the system.
(b) What is the direction of the electric field at every point on this surface?
(a) Identify an equipotential surface of the system.
(b) What is the direction of the electric field at every point on this surface?
Solution:
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(a) The system represents an electric dipole of charge strength $2 \mu \mathrm{C}$, electric dipole length $0.06 \mathrm{~m}$ and electric dipole moment
$=0.12 \times 10^{-6} \mathrm{C}-\mathrm{m}$.
By formula, $\mathrm{V}=\frac{1}{4 \pi \varepsilon_0} \frac{\mathrm{p} \cos \theta}{\mathrm{r}^2}$
(i) For points on axial line, $\theta=0^{\circ}$,
$\cos \theta=1$.
$\mathrm{V} \propto 1 / \mathrm{r}^2$.
The equipotential surfaces are spherical with centre at the centre of the dipole.
(ii) For points on equatorial line, $\theta=90^{\circ}, \cos \theta=$ zero.
The equatorial plane is plane of zero potential.
(b) Direction of electric field at every point of the equipotential surface is normal to the surface.
$=0.12 \times 10^{-6} \mathrm{C}-\mathrm{m}$.
By formula, $\mathrm{V}=\frac{1}{4 \pi \varepsilon_0} \frac{\mathrm{p} \cos \theta}{\mathrm{r}^2}$
(i) For points on axial line, $\theta=0^{\circ}$,
$\cos \theta=1$.
$\mathrm{V} \propto 1 / \mathrm{r}^2$.
The equipotential surfaces are spherical with centre at the centre of the dipole.
(ii) For points on equatorial line, $\theta=90^{\circ}, \cos \theta=$ zero.
The equatorial plane is plane of zero potential.
(b) Direction of electric field at every point of the equipotential surface is normal to the surface.
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