Problem Statement:

Two electric charges of 9 μC and -3 μC are placed 0.16 m apart in air. There are two points A and B on the line joining the two charges at a distance of 0.04 m from 9 μC and between the charges. Find the electric field at points A and B.

Solution:

Given:

Q1 = 9 μC

Q2 = -3 μC

r = 0.16 m

Distance of point A from Q1 = 0.04 m

Distance of point B from Q1 = 0.12 m

Distance of point B from Q2 = 0.04 m


Electric Field at Point A:

We know that the electric field at point A due to Q1 is given by:

E1 = (1/4πε0) * (Q1/r^2)

where ε0 is the permittivity of free space (8.85 x 10^-12 C^2/Nm^2)

Substituting the values, we get:

E1 = (1/4πε0) * (9 x 10^-6 / (0.16)^2)

E1 = 3187.5 N/C


The electric field at point A due to Q2 is given by:

E2 = (1/4πε0) * (Q2/r^2)

Substituting the values, we get:

E2 = (1/4πε0) * (-3 x 10^-6 / (0.20)^2)

E2 = -1687.5 N/C


The net electric field at point A is the vector sum of the electric fields due to Q1 and Q2:

E = E1 + E2

E = 3187.5 - 1687.5

E = 1500 N/C


Therefore, the electric field at point A is 1500 N/C.

Electric Field at Point B:

The electric field at point B due to Q1 is given by:

E1 = (1/4πε0) * (Q1/r^2)

Substituting the values, we get:

E1 = (1/4πε0) * (9 x 10^-6 / (0.12)^2)

E1 = 5208.33 N/C


The electric field at point B due to Q2 is given by:

E2 = (1/4πε0) * (Q2/r^2)

Substituting the values, we get:

E2 = (1/4πε0) * (-3 x 10^-6 / (0.04)^2)

E2 = -70685.0 N/C


The net electric field at point B is the vector sum of the electric fields due to Q1 and Q2:

E = E1 + E2

E = 5208.33 - 70685.0

E = -