Electric Potential Homework: Kinetic Energy Calculation

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The discussion revolves around calculating the kinetic energy of an alpha particle released between two parallel plates with a potential difference of 529 V. The key equation used is U = qV, where U represents electric potential energy, q is the charge of the alpha particle, and V is the potential difference. Participants clarify that gravitational potential energy is not relevant in this scenario since the problem focuses solely on electric potential energy. The electric field (E) is derived from the potential difference and distance between the plates, but it is emphasized that the potential difference alone suffices for the calculation. The conversation concludes with participants confirming their understanding of the problem.
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Homework Statement


The potential difference between two parallel plates is 529 V. An alpha particle with mass of 6.64 kg and charge of 3.20 C is released from rest near the positive plate. What is the kinetic energy of the alpha particle when it reaches the other plate? The distance between the plates is 19.4 cm.

Homework Equations


U_{initial}=K_{Final}
U=q_{0}E\DeltaR

The Attempt at a Solution


So I am trying to find E now so I can plug that in for U. E=V/distance right? so
E= 1058/.194?
 
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You can calculate the electric potential energy immediately. You know q, you know V...

the question is do you need to consider gravitational potential energy for this problem... are the capacitor plates up and down, or side to side?
 
Last edited:
No, it is simply taking into account the electric potenial energy, no gravity included in the problem.
U=qEr(distance) so don't I need to calculate E?
 
Winzer said:
No, it is simply taking into account the electric potenial energy, no gravity included in the problem.
U=qEr(distance) so don't I need to calculate E?

You already know V. So U = qV.
 
V being the potenial difference between the two plate -> 529V correct?
 
Winzer said:
V being the potenial difference between the two plate -> 529V correct?

yeah.
 
ok I got it thanks
 
Winzer said:
ok I got it thanks

no prob.
 
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