Unknown Charge on Rectangle?

A charge is to be placed at the empty corner to make the net force at corner A point along the vertical direction. What charge (magnitude and algebraic sign) must be placed at the empty corner if the three charges have the same charge of +8.45 μC?

Force = kq1q1/r^2

where k = 8.99 x 10^6

Attempt:

So we know that the horizontal charges must cancel out since the net force is vertical..ok..so we know there is a downwards vertical force on A coming from the top right charge, a leftwards horizontal charge from A coming from the bottom right charge...I know the unknown charge on A (at an angle of 14 degrees [tan theta = d/4d]) will be diagonal...not too sure where to go from here!! Any help would be appreciated!!

It is not a relevant question to mathematics and it is a physic problem. However, if you name the other angles of the rectangle as $B,C$ and $D$, as is shown in the following picture, then it is apparent that the force exerted on $A$ by $B$ is vertically downward and the force exerted on $A$ by $D$ is horizontally and toward left. Hence we should cancel the effect of the force Exerted by D, and we do not care about the force due to the charge at angle $B$, since it is vertically, as we want.

So we must choose the charge at point $C$, such that its horizontal component of its force on $A$ cancels the force due to charge $D$. So its horizontal component must be toward right, and hence the sign of the charge is opposite to the sign of charge at $A$. Hence it has a negative charge.

Let $F_{MN}$ denots the force exerted on $M$ by $N$, then we must have:

$F_{AC}\times \cos{\theta}=F_{AD}\Rightarrow k\frac{q_Cq_A}{17d^2}\times \cos{\theta}=k\frac{q_Dq_A}{16d^2}$

in which $\cos{\theta}=\frac{4}{\sqrt{17}}$. Hence $q_C=\frac{17\sqrt{17}}{64}q_D=9.25 \mu C$

• Hey math and physics, we're all unified in nature!! This was the correct answer!! thanks so much! But where did you get the 17 from? Oct 12 '14 at 23:19
• 4*4+1=17, the diagonal length of triangle Oct 12 '14 at 23:27
• but isn't the length of the diagonal root(17d), therefore when you square it, doesnt it just give you 17d? not 17d^2? Oct 12 '14 at 23:32
• yes, it is. but there is a d at nominator and another d in denominator which cancel each other, and therefore their ratio is 4/sqrt(17) Oct 12 '14 at 23:33