Question

1. A particle, Q, with a positive charge of 2.5 nC is isolated from all other charged objects. Point A is 1.5 cm to the right of the charge. Point B is 1.5 cm below the charge. Point C is 2.5 cm to the left of the charge.

a. What is the electric potential at points A, B, and C?

b. What is the potential energy of an electron at each point?

c. If an electron is placed at rest at point C, will it move? If so, in what direction? If not, why?

Answer 1

Given

Q = 2.5 nc

the points A at 1.5 cm to the right of the charge Q

the points B at 1.5 cm belowthe charge Q and

the points C at 2.5 cm to the leftt of the charge Q

we know that the potential is V = k*Q/r

V_A = 9*10^9*2.5*10^-9/(1.5*10^-2) = 1500 V

V_B = 9*10^9*2.5*10^-9/(1.5*10^-2) = 1500 V

V_C = 9*10^9*2.5*10^-9/(2.5*10^-2) = 900 V

b. potential energy of electron is  

U = kq1*q2/r

U_A = v_A*q = 1500*1.6*10^-19 C J = 2.4*10^-16 J

U_B = v_B*q = 1500*1.6*10^-19 C J = 2.4*10^-16 J

U_C = v_C*q = 900*1.6*10^-19 C J = 1.44*10^-16 J

c.

if the charge q electron is placed at point c then  

the charge Q attracts electron so that it will move in the +ve x direction

From coulomb's law  

F = kQ*q/r^2

F = -9*10^9*2.5*10^-9*1.6*10^-19 /(2.5*10^-2)^2 N

F = -5.76*10^-15 N

the -ve sign indicates the attractive force so the direction is to the +ve x direction