electric field?

Question

an electric field is given by Ex = (2kN/C)X^3. Find the potential difference between the points on teh x-axis at x = 1m and x= 6m. Answer in kV.

Answer 1

For an elemental space, potential difference is given by
dV = - E dx

variable field is E = 2k x^3 (N/C) here k is constant having dimensions of N/(C*meter^3) and x is distance in meter

The PD between two points is
x2..........x2...........6
∫ dV = - ∫ E dx = - ∫ 2k x^3 dx
x1.......x1............1
.....................................6
V1 - V2 = + 2k [ x^4 / 4] [(N) m^4/{C*m^3}]
.....................................1

V1 - V2 = + (1/2) k [ (6)^4 -1] (N-M/C) or volt
= 647.5 (k) volts
answer

Answer 2

Integrate the field between the two points.

Answer 3

potential difference = - integral E.dx =-2k x (6-1)= 10 kV