# Figure below shows four charges at the corners of a square of side L.What magnitude and sign of charge Q will make the force on charge q 0?

Fx = F1 + F2 + F3

F1 = k(-10nC * q) / L^2

F2 = 0 (No force in the x direction)

F3 = k(Q * q)cos(45) /2 L^2

Fx = kq/L^2 (-10nC + Qsqrt(2)/4)

Fy = F1 + F2 + F3

F1 = 0

F2 = k(-10nC*q)/L^2

F3 = k(Q * q)sin(45) /2 L^2

Fy = kq/L^2 (-10nC + Qsqrt(2)/4)

Fnet = sqrt(Fx^2 + Fy^2) = 0

Removing the common kq/L^2

2*(100nc + Q^2*2/16 - 5Qsqrt(2))

= 200nC + Q^2/4 - 10Qsqrt(2)

Using the quadratic formula, I am getting

Q = 20 sqrt(2) nC

Is this correct, please let me know if I have made a mistake. Thank you.

the Quadratic does not have a - answer unless the "b" second term of the quadratic is positive.

### 1 Answer

- NCSLv 73 months agoFavorite Answer
You are correct, but you've made it harder than you needed to. No quadratic is necessary.

While it is true that

Fnet = sqrt(Fx^2 + Fy^2) = 0

it also must be true that

Fx = Fy = 0.

Using your

Fx = kq/L^2 (-10nC + Qsqrt(2)/4)

and setting it to zero, I proceed with

-10nC + Qsqrt(2)/4 = 0

and then

Q = 28 nC

If you proceed with Fy instead you get the same result.

I typically set it up a little more simply and consider that the magnitude of the field on q due to Q must equal the magnitude of the field due to the other two corner charges:

kQ / (2L²) = 2*k*10nC*cos45º / L²

which leads directly to

Q = 4*cos45º*10nC

and your result.

Hope this helps!

You're welcome. Thanks for the BA!