The magnitude E of an electric field depends on the radial d

The magnitude E of an electric field depends on the radial distance r according to E = A/r^4, where A is a constanttt with unit volt-cubic meter. As a multiple of A, what is the magnitude of the electric potential difference between r_1 = 3.91 m and r_2 = 8.49 m? Number Units the tolerance is +/-9percentagee

Solution

the electric field is given by,

E = A/r⁴

thus, the potential difference is,

    V = - A/r⁴ dr

          = -A [r^-4+1 / -4+1]   limits: r1 = 3.91 m and r2 = 8.49 m

          = -(A/3)[-1/r³]           limits: r1 = 3.91 m and r2 = 8.49 m

          = -(A/3)[-{(1/8.49)³ - (1/3.91)³]

          = -(5.03X10^-3)A

Magnitude: V = (5.03X10^-3)A units: V