Location where charge is cancelled out

[scholio]

Homework Statement

two charges, are +10 microcoulombs and +6microcoulombs are placed 40cm apart. find the loaction along the line joining the two charges when the force on a -2microcoulomb chrage would be zero. consider only locations between the two positive charges

Homework Equations

E = E1 + E2 = (kqa/r^2) + (kqa/r^2)... where E is electric field, and E1 is electric field of one +charge qa, and E2 is the electric field of the other charge qb

k = 9*10^9 constant


E = F/q where F = force, q is charge = -2microcoulombs, and E is from the first equation

The Attempt at a Solution

i am not sure if i am using the correct equations

first i used the frst eq and solved for E1 and E2 using the k and the given charges and radius r = 40cm =0.4m, i summed E! and E2 to get E

then used the second equation, and subbed in E from the first eq in for E in the second, used q = -2microcoulombs for q, but the thing is i am trying to find where the force is zero so, if i sub in 0 for F in the second equation i get no where

i need an equation that allows me to factor in force and the q= -2microcoulombs charge. i need to find r

help appreciated..

 

[G01]

First things first:

In your first equation you have two different r's:

$$E=E_1+E_2=\frac{kq_a}{r_a^2}+\frac{kq_b}{r_b^2}$$

where r_a and r_b are the distance between the position we care about and chareg a or b, respectively.

Now, since F=qE and we know the charge isn't zero, the point where the force is zero has to be the point where the field, E, is zero. Thus, the point we care about, call it x, is where the field is zero.

HINT:

Can you get x involved in the first equation for the total E-field?

If so, you should be able to solve for x, knowing the field at x is zero.

 

[scholio]

this is what i thought of but it just doesn't seem right at all

E = kq_c/r^2 where k = 9*10^9, q_c = -2 microcoulombs, r = (0.4 - x)

other that i don't what else, hint...?

 

[MarkusNaslund19]

You need to write r_a and r_b in terms of x. Then you can solve G01's equation for x.