Question

Some equipotential surface is shown in the figure. What can you say about the magnitude and the direction of the electric field? 

Answer 1

(a) The electric field is always perpendicular to the equipotential surface. (As shown in the figure)  

So, the angle between \[\vec{E} \text{ and } \vec{dx}\] = \[90^\circ+ 30^\circ\]

Change in potential in the first and second equipotential surfaces, dV = 10 V 
so,

\[\vec{E} . \vec{dx} = - \] dV

\[\Rightarrow \text{ Edx } \cos(90^\circ+ 30^\circ) = - \] dV
\[ \Rightarrow E(10 \times {10}^{- 2} )\cos120^\circ= - 10\]
\[ \Rightarrow E = 200\] V/m

The electric field is making an angle of 120°  with the x axis. 

(b) The electric field is always perpendicular to the equipotential surface. 

So, the angle between  \[\vec{E}\] and \[\vec{dr}\] = 0° 

Potential at point A,

\[V_A  = \frac{1}{4\pi \epsilon_0}\frac{q}{r} = \]60  V

\[\Rightarrow \frac{q}{4\pi \epsilon_0} = 60 \times r\]
\[ \Rightarrow \frac{q}{4\pi \epsilon_0} = 0 . 6\]

So, electric field,

 

\[E = \frac{q}{4\pi \epsilon_0} \times \frac{1}{r^2} = \frac{0 . 6}{r^2}\]

The electric field is radially outward, decreasing with increasing distance.