Question

Two charges, -24

Answer 1

1.9m -31muC ----------------------------------------... +13muC------------x

The x is where the charge will be, since it must be outside and relative to the positive charge. The distance from the +13muC to x is what we are solving for

The electric field is defined by what a positive charge would do in that place, so the other positive charge would push it to the right (likes repel) and the negative would pull it to the left (opposites attract).

Ok, thats where we do the math.

The electric field is defined as E = k q/r^2 where r is the distance between the charge and point and k is a constant (don't worry about it).

We want both the pull of the negative or its electric field vector to equal that of the positives.

So- kq/r^2 = kq/r^2 basically. Looking at the diagram, the distance the point is from the positive charge will be x distance, ok. So this means the electric fields will be kq/x^2 = kq/(r+x)^2 where r is now 2.2m.

This makes sense because the point must be farther away from the negative than the positive because its charge is a greater magnitude.

Ok, work it out - k (+18muC)/x^2 = k (-24muC)/(r+x)^2 (dont worry about k, it cancels; dont even worry about muC, because mu cancels, and ignore the +/-).

18/x^2 = 24/(2.2m + x)^2 (Now we have to solve).

Also, there is a possible negative answer, but we must rule it out becuase this would mean that the point is inside the two, where it can't be). 14.2 m relative to the positive charge.

(B). There will be a net force of 0 because at this point the electric field (net) is zero (that's what we just did). The unit for electric field most commonly is E = N/C. So to find force, multiply q x E = 26muC x 0 N/C = 0.000000000N.