The distance between the points A and B on two equipotential lines with V1=4 V and V2=3.1 V is 3.7 cm. What is the average electric field (in Volt/m) at the midpoint C expressed to one decimal place?

24.32 V/m

Explanation:

We are given that

The distance between the points A and B on two equipotential  lines=d=

By using

We have to find the average electric field in (volt/m) at the mid-point C expressed to one decimal places.

We know that the electric field=

Using the formula

Electric field=E=

electrical potential energy =0.072 j

explanation:

electrical potential energy is given by

u= q δv

q= charge=6 c

δv= potential difference and is given by

δv= e d

e= electric field

d= distance=5 mm=0.005 m

δv=2.4 (0.005)=0.012 v

now electric potential energy= u= q δv

u=6 (0.012)

u=0.072 j