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× 10

^{9}Nm

^{2}C

^{-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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