Given the distance between 7v equipotential and the 5v at point b is .9cm and the distance between the 3v equipotential and the 5v at point b is 1.0cm, what is the magnitude of the electric field at point b?
(a) 2.0 v/cm
(b) 2.2 v/cm
(c) 2.1 v/cm
(d) .5 v/cm

Electric field, E = 2.22 V/cm

Explanation:

Given that,

The distance between 7 V equipotential and 5 V at point B is 0.9 cm. The relation between electric field and electric potential is given by :

E = 2.22 V/cm

So, the magnitude of the electric field at point B is 2.22 V/cm. Hence, this is the required solution.

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