Anonymous
Anonymous asked in Science & MathematicsPhysics · 3 months ago

# 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.

Update:

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

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

Q = 4*cos45º*10nC

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