Two point charges of magnitude $+q$ and $-q$ are placed at $\left( { - \frac{d}{2},0,0} \right)$ and $\left( {\frac{d}{2},0,0} \right)$, respectively. Find the equation of the equipotential surface where the potential is zero.
Let the required plane lies at a distance $x$ from the origin as shown in figure.
Potential at point $P,$
$\frac{k q}{\left[\left(x+\frac{d}{2}\right)^{2}+h^{2}\right]^{1 / 2}}-\frac{k q}{\left[\left(x-\frac{d}{2}\right)^{2}+h^{2}\right]^{1 / 2}}=0$
$\therefore \frac{1}{\left[\left(x+\frac{d}{2}\right)^{2}+h^{2}\right]^{1 / 2}}=\frac{1}{\left[\left(x-\frac{d}{2}\right)^{2}+h^{2}\right]^{1 / 2}}$
$\therefore\left(x-\frac{d}{2}\right)^{2}+h^{2}=\left(x+\frac{d}{2}\right)^{2}+h^{2}$
$\therefore x^{2}-x d+\frac{d^{2}}{4}=x^{2}+x d+\frac{d^{2}}{4}$
$\therefore 0=2 x d$
$\therefore x=0$
The equation of the required plane is $x=0$ means $y z$ plane.
Assertion : Two equipotential surfaces cannot cut each other.
Reason : Two equipotential surfaces are parallel to each other.
Draw an equipotential surface for dipole.
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
A point charge $+Q$ is placed just outside an imaginary hemispherical surface of radius $R$ as shown in the figure. Which of the following statements is/are correct?
(IMAGE)
$[A]$ The electric flux passing through the curved surface of the hemisphere is $-\frac{\mathrm{Q}}{2 \varepsilon_0}\left(1-\frac{1}{\sqrt{2}}\right)$
$[B]$ Total flux through the curved and the flat surfaces is $\frac{Q}{\varepsilon_0}$
$[C]$ The component of the electric field normal to the flat surface is constant over the surface
$[D]$ The circumference of the flat surface is an equipotential
Define an equipotential surface.