Where Is the Electric Field Zero Between Two Charges?

[phy43]

Homework Statement

a 6uC + charge is at (0,0) and a 1uC + charge is at (0,1m). Where between them is the electric field equal to 0?

Homework Equations

F=qE
F=K(q_1)(q_2)/r^2

The Attempt at a Solution

F=K(q_1)(q_2)/r^2
F=(9.0*10^9)(1*10^-6)(6*10^-6)/(1^2)
F=5.4*10^-5 (The force each exerts upon each other)

F=qE
Since E=0, F=0

F=0 at:
0=K(q_1)(q_2)/r^2
Find r

0=(9.0*10^9)(1*10^-6)(6*10^-6)/r^2

And this is where we have a problem

 

[Doc Al]

You want to study the field from the two charges at various points, not the force between the two charges. Write an expression for the field at point X from each charge. Add to get the total field. (Careful with signs.)

 

[phy43]

Electric field at distance r for each charge
E=Kq/r^2

E_1=(9.0*10^9)(6*10^-9)/r^2
E_1 = 54/r^2

E_2=(9.0*10^9)(1*10^-9)/r^2
E_2 = 9/r^2

E_net = E_1+E_2 = 54/r^2 + 9/r^2
E_net = (54+9)(1/r^2)
E_net = 63/r^2
0 = 63/r^2

 

[Doc Al]

Electric field at distance r for each charge
E=Kq/r^2

If a point is a distance X from charge #1, how far is it from charge #2?

E_1=(9.0*10^9)(6*10^-9)/r^2
E_1 = 54/r^2

Instead of calling the distance r, call it X. Don't plug in numbers right away. (And μC = 10^-6, not 10^-9.)

What direction is the field?

E_2=(9.0*10^9)(1*10^-9)/r^2
E_2 = 9/r^2

Instead of r, write the distance in terms of X and the distance between the charges.

What direction is the field?

Note that different directions will get different signs.

 

[phy43]

Thanks, figured it out.