Charge given to the electroscope

Question

The question is: What is the charge given to the electroscope? Also each ball has a weight of 25g.

Here's how I started. First I draw the scheme on the forces acting one one ball (I took the one on the left).

Now I write the equations along the x and y axis.

\begin{cases} Tcos(60°)-F_{21}, & \text{along x} \\ Tsin(60°)-mg, & \text{along y} \end{cases}

So $T = \frac{mg}{sin(60°)}$

and (using Coulomb's law) $F_{21} = \frac{1}{4 \pi \epsilon_{0}} * \frac{(Q/2)²}{(2*60*sin(30°))²}$

Now I just isolate $Q$ so I end up with:

$ F_{21} = Tcos(60°)\\ \Leftrightarrow F_{21} = mg*tan^{-1}(60°)\\ \Leftrightarrow \frac{1}{4 \pi \epsilon_{0}} * \frac{(Q/2)²}{(2*60*sin(30°))²} = mg*tan^{-1}(60°)\\ \Leftrightarrow 9.0*10^9 * \frac{(Q/2)²}{(2*60*sin(30°))²} = mg*tan^{-1}(60°)\\ \Leftrightarrow \frac{Q^2}{4} = \frac{mg*tan^{-1}(60°)*(2*60*sin(30°))²}{9.0*10^9}\\ \Leftrightarrow Q^2 = \frac{4*mg*tan^{-1}(60°)*(2*60*sin(30°))²}{9.0*10^9}\\ \Leftrightarrow Q = \sqrt{\frac{4*25*10^{-3}*9.81*tan^{-1}(60°)*(2*60*sin(30°))²}{9.0*10^9}}\\ $

Plugin this into wolfram alpha gives : $5.632×10^{-4}$ but the answer in my book is $4.8×10^{-6} C$ (and there is only the answer, no explanations).

I tried to see where is my mistake but I really don't know so where is my error?

Answer 1

You have made two mistakes (highlighted in red below):-

$$\Leftrightarrow Q = \sqrt{\frac{4*25*10^{-3}*9.81*\color{red}{tan^{-1}(60°)}*(2*\color{red}{60}*sin(30°))²}{9.0*10^9}}\\$$

1. Instead of arctan, it should be $(\tan 60^{\circ})^{-1}$ i.e. tangent of the angle raised to the power of $-1$
2. The distance should be in metres (SI units) instead of centimetres, i.e. should be $0.6$ instead of $60$.

The corrected equation should be:- $$\begin{align}Q &= \sqrt{\frac{4*25*10^{-3}*9.81*\color{red}{tan(60°)^{-1}}*(2*\color{red}{0.6}*sin(30°))²}{9.0*10^9}}\\&=4.7598\times10^{-6} C\sim4.8\times10^{-6}C\end{align}$$