Question

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

Answer 1

Answer:

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 :

$E=\dfrac{\Delta V}{d}$

$E=\dfrac{7\ V-5\ V}{0.9\ cm}$

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.