Question

The magnitude E of an electric field depends on the radial distance r according to E = A/r4, where A is a constant with unit volt-cubic meter. As a multiple of A, what is the magnitude of the electric potential difference between r1 = 1.71 m and r2 = 2.89 m?

Answer 1

`\Delta V = 0.053 A`

Step-by-step explanation:

Electric field in a given region is given by equation

`E = (A)/(r^4)`

as we know the relation between electric field and potential difference is given as

`\Delta V = -\int E. dr`

so here we have

`\Delta V = - \int ((A)/(r^4)) .dr`

`\Delta V = (A)/(3r_1^3) - (A)/(3r_2^3)`

here we know that

`r_1 = 1.71 m` and
`r_2 = 2.89 m`

so we will have

`\Delta V = (A)/(3)((1)/(1.71^3) - (1)/(2.89^3))`

so we will have

`\Delta V = 0.053 A`