Question

Two charges of equal magnitude and opposite sign form a dipole. Let q₁ = 4 nC and q₂ = -­ 4 nC, and the charges are located on the y axis at y = 1.0 mm and y = ­-1.0 mm, respectively

A second dipole comprised of q_​3 = -­ 4 nC located at (0.002,0.001,0)m and q_​2 =­ -4 nC located at (0.002,­0.001,0)m is added to the dipole in #1.

d. What is the electric field of the two dipole system at the location (1,0,0)m? {N/C}

e. What is the electric field of the two dipole system at the location (4,0,0)m? {N/C}

f. Compare the values you find for part e and d. How is the answer to e related to d?

Answer 1

Note : Your problem seems to be wrong. So I am assuming following:

1st Diploe : q1 = 4nC at ( 0, 1X10^(-3) , 0)m q2 = -4nC at ( 0, - 1X10^(-3),0)

2nd Dipole : q3 = -4nc at ( 2X10^(-3),1X10^(-3),0 ) m q4 = 4nC at (2X20^(-3), 1X10^(-3) ,0 )

Field point ( 1,0,0)

Field at (1,0,0)m due to dipole 1 = k(2a) q/ r^3 , where k = 9.0 X 10^9 N-m^2/C^2 , a = 1X 10^(-3) , r= 1m

= 9.0X10^(9) Nt-m^2/C^2 X 4 X 10^(-9) C X 2 X 1 X 10^(-3) mrs/ (1m)^3

= 7.20X 10^(-2) N/c along negative Y-direction

Field at (1,0,0 ) due to Dipole 2 = 9.0X10^(9) Nt-m^2/C^2 X 4 X 10^(-9) C X 2 X 1 X 10^(-3) mrs/ (1-0.002)^3 m^3

= 7.24 X 10^(-2) N/C along positive Y-direction

Net Field in +Y-direction = 0.04 X 10^(-2) N/C = 0.0004 N/C

f.

Field required at (4,0,0) point

Field at (4,0,0)m due to dipole 1 = k(2a) q/ r^3 , where k = 9.0 X 10^9 N-m^2/C^2 , a = 1X 10^(-3) , r= 4m

= 9.0X10^(9) Nt-m^2/C^2 X 4 X 10^(-9) C X 2 X 1 X 10^(-3) mrs/ (4m)^3

= 0.1125 X 10^(-2) N/C along +Y-direction

Field at (4,0,0)m due to dipole 2 = k(2a) q/ r^3 , where k = 9.0 X 10^9 N-m^2/C^2 , a = (4-0.002)^3X 10^(-3) , r= 1m

E = 9.0X10^(9) Nt-m^2/C^2 X 4 X 10^(-9) C X 2 X 1 X 10^(-3) mrs/ (4-0.002)^3m)^3

= 0.1127 X 10^(-2) N/C along -Y direction

Net field along -Y direction = 0.0002 X 10^(-2) N/C = 0.000002 N/C

The field in (d ) part should be 64 times more intense than field in (e) Part . In fact as distance from origin increases the net field becomes close to zero , since effect of the two dipoles together will be like there being a zero charge at origin.