# Some equipotentia

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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