Question

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?

Answer 1

Answer:

24.32 V/m

Explanation:

We are given that

$V_1=4 V$

$V_2=3.1 V$

The distance between the points A and B on two equipotential  lines=d=$x_2-x_1=3.7 cm=3.7\times 10^{-2} m$

By using $1 cm=10^{-2}m$

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=$E=-\frac{dV}{dx}=-\frac{V_2-V_1}{x_2-x_1}$

Using the formula

Electric field=E=$-\frac{(3.1-4)}{3.7\times 10^{-2}}$

$E=24.32 V/m$