- #36
Saitama
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ehild said:The area is correct. Show details of getting ωf. The formula for vpf in #11 was still correct.
ehild
I had $$v_{pf}^2=\frac{kq^2(3\sqrt{2}-4)}{10mL_0} \Rightarrow \omega_f^2L_0^2/2=\frac{kq^2(3\sqrt{2}-4)}{10mL_0}$$
It is given in the question that ##kq^2/(mL_0^3)=10^4 s^{-2}##, plugging this, I reached the value of ##\omega_f##.