Question

Given two 2.00μC charges on the horizontal axis are positioned at x=0.8m, and the

other at x=-0.8m, and a test charge q = 1.28x10-18C at the origin.

(a) What is the net force exerted on q by the two 2.00μC charges? [5]

(b) What is the electric field at the origin due to 2.00μC charges? [5]

(c) What is the electric potential at the origin due to the two 2.00μC charges

Answer 1

Answer:Question 1

Given q1=2µC

q2=2µC

q= 1.2×10^-18C at origin

Net force exerted by two charges on q

F_1= force on q due to q1

F_2= force on q due to q2

F_net= F_(1-) F_2

= (Kqq_1)/r^2 - (kqq_2)/r^2 Then q_1=q_2=〖2×10〗^(-6)

F_net=0N

b) The electric field at charge q

E_net= E_1- E_2

= (kq_1)/r^2 - (kq_2)/r^2

Then q_1=q_2

E)_net= 0 N/C

c) The electric potential at origin due to two charge

V_net= V_(1 )- V_2

= (kq_1)/r - (kq_2)/r

Then q1= q2

V_net= 0 V

Step-by-step explanation: