Question

An electron is at the origin. (a) Calculate the electric potential V_Aat point A, x= 0.250 cm. (b) Calculate the electric potential V_Bat point B, x= 0.750 cm. What is the potential difference V_B- V_A? (c) Would a negatively charged particle placed at point Anecessarily go through this same potential difference upon reaching point B? Explain.

Answer 1

We have formula for electric potential,V = (k * q) / x -------(1)

where q = charge on an electron= 1.6 * 10^-19 C [in magnitude]

k = constant = 9 * 10^9

x = distance of the point from origin

a) Distance of point A from origin , x = 0.250 cm = 0.250 *10^-2 m [since 1 cm = 10^-2 m]

Thus from formula 1
Electric potential at A, VA= (k *q) / x
=(9 * 10^9 * 1.6 * 10^-19) / (0.250 *10^-2)
=5.760 * 10^-7 V [Answer part a]

b)At point B ,x = 0.750 cm=9.2 * 10^-2 m

Electric potential at B,VB = (k*q) / x
= (9 * 10^9 * 1.6 * 10^-19) / (0.750 * 10^-2)
=1.920 * 10^-7 V [Answer part a]

Potential difference,VA-VB= (5.760 * 10^-7 - 1.920 * 10^-7) V
=3.840 * 10^-7 V [Answer part b]

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.

[Answer part c]