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(a) Identify an equipotential surface of the system.
(b) What is the direction of the electric field at every point on this surface?
Asked by Kartika Last Modified
Answer 
Apoorva Purohit
(a) The situation is represented in the given figure.
An equipotential surface is the plane on which total potential is zero everywhere. This plane is normal to line AB. The plane is located at the mid-point of line AB because the magnitude of charges is the same.
(b) The direction of the electric field at every point on this surface is normal to the plane in the direction of AB.
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Gourab Saha
Clear Your Concepts and Score Higher
(a) As two charges are of opposite but of equal value, so equipotential surface is the perpendicular surface to the electric lines of forces that emanates from +2 μC charge and ends in -2 μC charge.
(b) the electric field is perpendicular to the equipotential surface and directed from +2 μC to -2 μC.
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(a) an equipotential surface is a plane on which potential is zero everywhere.
This plane is normal to the plane in which two point charges are placed.
Let normal plane be placed at C of x distance from +2μC charge.
Then, potential due to 2μC + potential due to -2μC = 0
Kq/x + k(-q)/(6 - x) = 0
x = 3 cm
Hence, the equipotential surface of the system appears at midpoint of the system of two given charges, and it is normal to the plane of charges.
(B) we know, the direction of the electric field from positive to negative charge.
So, the direction of the electric field at every point on the surface is normal to the plane in the direction AB.
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