Electrons enter charged capacitor

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Electrons entering a charged capacitor experience a force due to the electric field created by the capacitor's charge. This force acts on the electrons, causing them to accelerate in a specific direction, which is determined by the polarity of the capacitor. While the Earth's gravitational field is constant, the forces acting on the electrons and their resulting motion differ from gravitational effects. Understanding the relationship between electric fields and the motion of charges is crucial for grasping capacitor behavior. Clarifying these concepts can enhance comprehension of how capacitors function in electrical circuits.
Xuran Wu
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Homework Statement
Two electrons enter a region between charged capacitor plates with equal speed v Electron A is directed horizontally to the left while electron B is directed at 30 degrees below the horizontal. Each electron makes it to the left-hand plate. Which one of the following choices best compares the speeds of the charges(vA, vB)upon arrival at the left plate? Consider only the electrons A and B’s interactions with the constant field between the plates, ignoring any relativistic effects.
A. vA>vB
B. vA=vB
C.vA<vB
D. The answer depends on the size of the plate separation,d.
E. The answer depends on the magnitude of the charge, Q, on each plate
Relevant Equations
C=QV
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I cannot understand. I think that the constant electric fields and the direction the electrons ahead of cannot change the speed, but I can’t be sure about that. I really hope that someone can help me on that, thank you very mouch.
 
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Is there a force acting on each charge? If so, what is it and in what direction? Note that the Earth's gravitational field near the Earth's surface is also constant. Can you draw a parallel between the two situations?
 
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Thread 'Chain falling out of a horizontal tube onto a table'
My attempt: Initial total M.E = PE of hanging part + PE of part of chain in the tube. I've considered the table as to be at zero of PE. PE of hanging part = ##\frac{1}{2} \frac{m}{l}gh^{2}##. PE of part in the tube = ##\frac{m}{l}(l - h)gh##. Final ME = ##\frac{1}{2}\frac{m}{l}gh^{2}## + ##\frac{1}{2}\frac{m}{l}hv^{2}##. Since Initial ME = Final ME. Therefore, ##\frac{1}{2}\frac{m}{l}hv^{2}## = ##\frac{m}{l}(l-h)gh##. Solving this gives: ## v = \sqrt{2g(l-h)}##. But the answer in the book...

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