Question

The magnitude of electric field in a region varies with the distance r
E=10r +5
By how much does the electric potential increases in moving from point at r=1 to a point at r=10m.

Answer 1

E=10r+5

V= - integral(10r+5)dr= 5r^2+5r

At r=1

V= -10

At r=10

V= -550

Change =-540V

Potential should decrease not increase , as v proportional to 1/r

Answer 2

The electric potential increases in moving from point at r=1 m to a point at r=10 m is 540 Volts .

Given :

The magnitude of electric field in a region varies with the distance r  E=10r +5 .

We know potential difference between two point A and B is given by :

$V_a-V_b=-\int_{a}^{b}E.dr$

Here a is point at 1 m and b is point at 10 m .

$V_a-V_b=-\int_{1}^{10}(10r+5).dr\\\\V_a-V_b=-_1^{10}[ 5r^2+5r]\\\\Vab=540\ V .$

The electric potential increases in moving from point at r=1 m to a point at r=10 m is 540 Volts .

Hence , this is the required solution .

Learn More :

Electric Field and Potential

https://brainly.in/question/15720812