Where Should a -3 MicroCoulomb Charge Be Placed to Achieve Zero Net Force?

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SUMMARY

The placement of a -3 microCoulomb charge to achieve zero net force between two charges of 4mC and 8mC, separated by 80 cm, is determined to be approximately 33.137 cm from the 4mC charge. The calculations involve using Coulomb's law, represented by the equation F = (k*q1*q2) / (r^2), to equate the electric fields produced by both charges. By setting the magnitudes of the electric fields equal and solving the resulting equations, the position of the -3 microC charge is accurately found.

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Homework Statement


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.

Homework Equations



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

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 doesn't seem right, what equation I am not using to get this answer?
 
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Because the net force acting on the charge is 0N, we know that the electric field is 0 N/C.

Now, through simple arguments 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 equation 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.
 
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
 
you now have a simple equation 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.
 
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.
 

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