Question

Assume the electric field has a magnitude of 59.0 ​m ​ ​V ​​ , and the particle has a net positive charge of Q=+5.40 C. Calculate the magnitude of the electric potential difference, \vert \Delta V \vert = \vert V_B - V_A \vert∣ΔV∣=∣V ​B ​​ −V ​A ​​ ∣.

Answer 1

V = 1.69 * 10^6 V

Step-by-step explanation:

Parameters given:

Electric field, E = 59V/m

Charge, q = 5.40C

We need to first find the distance between the electric charge and the point of consideration to be able to find the Electric potential difference.

Electric field is given as:

E = (kq/r^2)

k = Coulombs constant

=> r^2 = kq/E

=> r^2 = (9 * 10^9 * 5.4) / 59

r^2 = 8.2 * 10^8

r = 2.84 * 10^4 m

We can now find the Electric Potential by using:

V = kq/r

Hence,

V = (9 * 10^9 * 5.4) / (2.84 * 10^4)

V = 1.69 * 10^6 V