Recent content by mmfiizik

Thank you all for explanations. I tried to prove this to myself by contradiction, but it might be an oversimplification. If we assume that there is uniform charge density on both plates, it would mean that at every point out of that plate displacement field ##\text{D}## would be the same. Then...

But since the top slab doesn’t move, top plate's surface-charge density (or charge distribution) shouldn’t change. Then the normal displacement at the plate is the same everywhere (D=σ=const). Because the top layer is the same material ε_r1 across the whole width, the electric field there would...

I am working on part (f) of a exercise. The setup is a square parallel-plate capacitor with plate side a and separation d. Initially the gap is filled by two layers on top of each other. The upper half has relative permittivity epsilon_{r1} and the lower half has epsilon_{r2}. Then the lower...

I just don't understand should I take u relative to the plane or relative to the ground. I tried to solve it like this: $$p_{final}=m_{0}v-m(u-v)-M(u-v)$$ $$p_{initial}=m_{0}v$$ $$\Delta p=-m(u-v)-M(u-v)$$ ##m_0## is mass of the plane. $$F=\Delta p$$ $$F=-m(u-v)-M(u-v)=(m+M)(v-u)$$...

But in the task there is written biggest part of energy, not bigger. So there is only one ratio of masses with which ##m_1##'s KE is biggest, right? And I have to maximize for ##m_1## energy, because it's the top one.

Yes, I made some mistakes in previous solution. Expression for energy gain should be written for the first ball, not the second. $$E=\frac{m_{1}v_{1}^{2}}{m_{1}v_{0}^{2}+m_{2}v_{0}^{2}}$$ solving ##v_1## from (2) $$v_{1}=\frac{v_{0}(3m_{2}-_{}m_{1})}{m_{1}+m_{2}}$$ replacing ##v_1^2## gives...

Ok, so then ##-2v_0=v_2-v_1## replacing ##v_1## in (2) $$m_{2}{v}_{0}=m_{2}{v}_{2}+m_{1}{v}_{2}+3m_{1}{v}_{0} $$ $${v}_{2}=\frac{v_{0}(m_{2}-3m_{1})}{m_{1}+m_{2}}$$ Energy converted = first ball energy / total energy $$E=\frac{m_{2}v_{2}^{2}}{m_{1}v_{0}^{2}+m_{2}v_{0}^{2}}$$...

So then (2) should be $$m_{2}{v}_{0}-m_{1}{v}_{0}=m_{2}{v}_{2}+m_{1}{v}_{1}$$ then I get ##2v_0=v_1-v_2##, which gives me same result if I square it. I think somehow I should find maximum velocity for ##m_1## after first collision with ##m_2##. Then it would get maximum kinetic energy.

With given information in the problem I can't decide which direction balls will go after collision. I assumed they will go in opposite directions. I know that is not a full solution, but I can't come up with anything else. From conservation of energy we have...