Question

An electron is at the origin.

(a) Calculate the electric potential V_A at point A, x = 0.860 cm.
V

(b) Calculate the electric potential V_B at point B, x = 1.020 cm.
V

What is the potential difference V_B - V_A?
V

(c) Would a negatively charged particle placed at point A necessarily go through this same potential difference upon reaching point B? Explain

Answer 1

Charge of an electron,q=1.6x10^-19 C


a)Distance,x=0.860 cm=8.6x10^-3 m
Electric potential at A, VA=k q/x
=9x10^9x1.6x10^-19 /(8.6x10^-3)
=1.67x10^-7 V

b)Distance,x=1.020 cm=10.2x10^-3 m

Electric potential at B,VB=k q/x
=9x10^9x1.6x10^-19 /(10.2x10^-3)
=1.41x10^-7 V

Potential difference,VA-VB=(1.67x10^-7 -1.41x10^-7) V
=0.258x10^-7 V

c)It depends on the quantity of charge of that particle.If the particle carries equal charge then the potential difference will be the same.On the other hand, if the particle carries greater or smaller charge,the potential difference will be greater or smaller.