Recent content by msw1

Oh, that was it! So ##|Q_A|<|Q_B|## like the question asked, but ##Q_A>Q_B##, and so there is no contradiction. Thanks!

Sure, energy transferred thermally in the isobaric processes is $$Q=NC_p\Delta T$$ $$Q=N\left(\tfrac{d}{2}+1\right)k_b\Delta T$$ And ##\Delta T = \frac{P\Delta V}{Nk_B}## by the ideal gas law so $$Q=P\Delta V\left(\tfrac{d}{2}+1\right)$$ And then for the isochoric processes...

Here is the figure: The answer is $$Q_A<Q_B$$ which I can show by calculation using the above equations. What's confusing to me is I thought that the change in internal energy was a state function. Which would mean since the initial and final points are the same, $$\Delta E_A=\Delta E_B$$ or by...

Thanks for the feedback! I found my mistake. It was very subtle. The charge of particle 3 is given in microcoulombs, not nanocoulombs. Very annoying!

r_{13}=r_{23}=\sqrt{(30*10^{-3})^2+(90*10^{-3})^2}=\sqrt{9*10^{-3}}\\ F^E_{13}=F^E_{23}=9E9\cdot\frac{5*10^{-9}\cdot3*10^{-9}}{9*10^{-3}}=1.5*10^{-5}\\ \theta=tan^{-1}(\frac{90*10^{-3}}{30*10^{-3}})=71.565\,degrees\\ \vec{F}^E_{13}=<F^E_{13}cos\theta, F^E_{13}sin\theta> = <4.743*10^{-6}...