Question

explain clearly in detail​

Answer 1

Answer:

Given:

A voltage vs distance graph has been provided in the question

To find:

Electric Field intensity at distances

r = 1, 6, 9 metres.

Concept:

The relationship between voltage and electric Field intensity can be given as follows :

E = - δV/δr ,

where δV => change in Potential

δr => change in position and E => Electric Field Intensity.

Calculation:

1. You have to calculate the Potential at

r = 1m using similarity of triangles.

Let potential (at r = 1m) be V1

V1/10 = 1/2

=> V1 = 5 volts.........(1)

E = - δV/δr

=> E = -(5-0)/(1-0)

=> E = -5 V/m.....(at r=1m)

2. At r = 6 m,

there is no change in Potential as indicated by line parallel to x-axis.

so, δV = 0

Hence (- δV/δr) = 0

So Electric Field intensity = 0

(at r=6m)

3. At r = 9m

Again calculate the potential at r= 9m using similarity of triangles.

Let the potential be V3

V3/10 = 1/2

=> V3 = 5 volts.

So , E = - δV/δr

=> E = -(0-5)/(10-9)

=> E = +5 V/m

So the correct option is Option B)