Advertisement Remove all ads

Some Equipotential Surface is Shown in the Figure. What Can You Say About the Magnitude and the Direction of the Electric Field? - Physics

Numerical

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

Advertisement Remove all ads

Solution

(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. 

Concept: Electric Field - Introduction of Electric Field
  Is there an error in this question or solution?
Advertisement Remove all ads

APPEARS IN

HC Verma Class 11, Class 12 Concepts of Physics Vol. 2
Chapter 7 Electric Field and Potential
Q 62 | Page 123
Advertisement Remove all ads
Advertisement Remove all ads
Share
Notifications

View all notifications


      Forgot password?
View in app×