# Homework Help: College Physics 2 Electricity

1. Oct 22, 2008

### creeps228

1. The problem statement, all variables and given/known data
Two charges are separated as shown. Where would you place a -3 microCoulombs (mC) charge such that the net force on this charge is 0?

+ +
4mC 8mC
Distance between charges are 80cm apart.

2. Relevant equations

F= (k*q1*q2) / (r*r)

3. The attempt at a solution

I found out the Force on q1 ( i made q1 the 4mC charge and q2 the 8mC charger) to q2
it came to be 4.5*10^(-1) N. Same for the F for q2 on q1. Then i manipluated the equation to slove for r. Once that was established i but the -3mC charge in the equation with the 4mC charge and F being 4.5*10^(-1) giving me an approximation of .5m aka 50 cm. But that doesnt seem right, what equation im not using to get this answer?

2. Oct 23, 2008

### christensen

Because the net force acting on the charge is 0N, we know that the electric field is 0 N/C.

Now, through simple arguements we know the point we are looking for is inbetween the two charges and that the field strength from charge 1 has to equal the field strength from charge 2 (in magnitude).

*Let a be the distance from q1 to the point of E = 0 and b be the distance from q2 to the point of E = 0.

So, E1 = E2

k*q1/a2 = k*q2/b2

Cancel k to get,

q1/a2 = q2/b2

you can reorganize the equasion to be simplier to use later on to,

q1*b2 = q2*a2

Now you also have a + b = 80cm

By substituting into another, you can solve for a or b, and thus solve for the other, giving you the point in which there is zero force on the charge.

3. Oct 27, 2008

### creeps228

I subsituted b for 80-a

so using the equation you gave q1= 4mC q2=8mC

4mC*[(80-a)^2] = 8mC*[a^2]
= 4mC* (6400-160a+[a^2]) = 8mC*a^2

which seems more confusing then just doing problem itself

4. Oct 27, 2008

### christensen

you now have a simple equasion to solve for the value of "a"

4mC* (6400-160a+[a^2]) = 8mC*a^2
(6400-160a+[a^2]) = 2*a^2
6400 = a^2 + 160a

a = 33.137 (approx)

so the charge must be placed 33.137cm from the 4mC charge.

5. Oct 28, 2008

### creeps228

Thankyou. The force of the 3mC charge from the other two chargers were about .02 difference. Basically one was 9.7*10^(-11) and the other was 9.9*10^(-11). Close to zero as possible, but thankyou.