Q. 625.0( 5 Votes )

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


Answer :


Given:
From figure (a)
Angle between equipotential surfaces and the displacement x
:θ = 30°
Change in potential : dV = 10 V
Change in displacement between two consecutive equipotential surfaces: dx = 10 cm = 0.1 m
From figure (b)
Increase in radius from center: dr= 10cm=0.1m
Formula used:
(a)
As we know that the electric field is always perpendicular to the equipotential surface

From the above diagram, the angle between Electric field
and dx ; θ’ = 90° +30° =120°
Change in electric potential is given as:




Hence the magnitude of electric field is 200 V/m making an angle of 120° with the x axis.
(b)
As we know that the electric field
is always perpendicular to the equipotential surface

Radius increases by: dr= 10 cm = 0.1 m
As
is perpendicular to the equipotential surface, and dr would be along same line as shown in the figure above.
Thus angle between
and dr : θ = 0

We know that potential at a point due to a charge q is given as:
Where, k is a constant and k= =9× 109 Nm2C-2 . q is the point charge and r is radius of that surface.
Consider potential at point A where r=0.1 m


Electric field is given as:
Substituting value of kq we get,
Hence, the magnitude of the electric field is and it’s direction is radially outward with decreasing with increasing radii.


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