What is an equipotential surface ? Draw an equipotential surfaces for a
$(1)$ single point charge
$(2)$ charge $+ \mathrm{q}$ and $- \mathrm{q}$ at few distance (dipole)
$(3)$ two $+ \mathrm{q}$ charges at few distance
$(4)$ uniform electric field.
If the electric potential along every point of any imaginary surface in an electric field is the same, then such a surface is called an equipotential surface.
$(1)$ The electric potential of a single charge $q$ to distance $r$ is,
$\mathrm{V}=\frac{k q}{r} \quad \therefore \mathrm{V} \propto \frac{1}{r}$
This shows that $\mathrm{V}$ is a constant if $r$ is constant. Hence surface passing through points having same $r$ obtain as spherical and its radius is $r$ and $q$ is the electric charge on the centre as shown in below figure.
More than one equipotential surfaces can be drawn for different radius.
Clearly the field lines at every point is normal to the equipotential surface passing through that point.
The electric field lines for a single charge $q$ are radial lines starting from or ending at the charge are depending on whether $q$ is positive or negative which is shown in this figure.
The angle between the electric lines of force and the equipotential surface is
Thepoints resembling equal potentials are
Draw an equipotential surface for dipole.
Assertion $(A):$ A spherical equipotential surface is not possible for a point charge.
Reason $(R):$ A spherical equipotential surface is possible inside a spherical capacitor.
A uniformly charged solid sphere of radius $R$ has potential $V_0$ (measured with respect to $\infty$) on its surface. For this sphere the equipotential surfaces with potentials $\frac{{3{V_0}}}{2},\;\frac{{5{V_0}}}{4},\;\frac{{3{V_0}}}{4}$ and $\frac{{{V_0}}}{4}$ have rasius $R_1,R_2,R_3$ and $R_4$ respectively. Then