Show that the direction of electric field at a given is normal to the equipotential surface passing through that point.
If the electric field were not normal to the equipotential surface, it would have non-zero component along the surface.
To move a unit test charge against the direction of the component of the field work would have to be done.
But this is in contradiction to the definition of an equipotential surface because the potential difference between any two points on the surface is $\Delta \mathrm{V}$ is zero so in
$\therefore$ Work $\mathrm{W}=q \Delta \mathrm{V}, \Delta \mathrm{V}=0, \therefore \mathrm{W}=0$
and work done in electric field $\overrightarrow{\mathrm{E}}$ with displacement $\vec{d} l$
$\mathrm{W}=\overrightarrow{\mathrm{E}} \cdot \overrightarrow{d l}=\mathrm{E} d l \cos \theta$ $0=\mathrm{E} d l \cos \theta$ $\therefore 0=\cos \theta \quad[\because \overrightarrow{\mathrm{E}} \neq 0, \overrightarrow{d l} \neq 0]$ $\therefore \quad \theta =\frac{\pi}{2}$
`Hence, the electric field $\overrightarrow{\mathrm{E}}$ is normal to the equipotential surface at that point.
Two conducting hollow sphere of radius $R$ and $3R$ are found to have $Q$ charge on outer surface when both are connected with a long wire and $q'$ charge is kept at the centre of bigger sphere. Then which one is true
Write the characteristics of equipotential surface.
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.
Given below are two statements: one is labelled a
Assertion $(A)$ and the other is labelled as Reason$(R)$
$Assertion$ $(A)$ : Work done by electric field on moving a positive charge on an equipotential surface is always zero.
$Reason$ $(R)$ : Electric lines of forces are always perpendicular to equipotential surfaces.
In the light of the above statements, choose the most appropriate answer from the options given below
Thepoints resembling equal potentials are