Understanding Electric Force on Q in Relation to Distance d and Charge Q1

[adamaero]

Homework Statement

Three point charges Q, Q1, and Q2 are separated by a distance d from each other in a homogeneous medium. Find the electric force on Q.

Relevant Equations

Pythagorean theorem

x^2 + d^2/4 = d^2
x = sqrt(3d^2/4)
F_1x = sqrt(3)*d*k*Q1/2.
In the solution, where did the "d" in the numerator go??

Is my math wrong?

 

[haruspex]

F_1x = sqrt(3)*d*k*Q1/2.

I cannot make sense of that equation. I think you have some typos, and an unclear notation.
Is that $F_1x$ or $F_{1x}$ on the left?
Why are you multiplying by d instead of dividing by d²?

 

[Doc Al]

x^2 + d^2/4 = d^2
x = sqrt(3d^2/4)

Why are you calculating the distance "x"?

 

[adamaero]

I cannot make sense of that equation. I think you have some typos, and an unclear notation.
Is that $F_1x$ or $F_{1x}$ on the left?
Why are you multiplying by d instead of dividing by d²?

"F" sub "1x" the exact same in the solution except without the "d".
d² is in the term "k".

 

[adamaero]

Why are you calculating the distance "x"?

I thought that is for the direction of the vector defined by a₁ (in F₁ = k*Q₁*a₂).

 

[haruspex]

"F" sub "1x" the exact same in the solution except without the "d".
d² is in the term "k".

Hmm... yes I see that is how they have used k in the solution too, but it is very nonstandard. The usual is $k=\frac 1{4\pi\epsilon_0}$.

I thought that is for the direction of the vector defined by a₁ (in F₁ = k*Q₁*a₂).

A component is obtained by multiplying by the cosine of the angle. You need to divide by the hypotenuse.

 

[adamaero]

A component is obtained by multiplying by the cosine of the angle. You need to divide by the hypotenuse.

That makes sense where the "d"s cancel out...but I don't understand why the Pythagorean theorem can't be used alone.
cos(θ) = (d/2)/d
cos(θ) = 1/2

So it's the magnitude of "x"combined with the direction (defined by the cosine-term), and that's just what makes a vector...and not a scalar?

 

[haruspex]

So it's the magnitude of "x"combined with the direction (defined by the cosine-term),

No, it's the magnitude of the force multiplied by the cosine. The only relevance of the magnitude of x is in finding the value of the cosine, as x/d (which is √3/2, not 1/2).