Formula for electric potential energy

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In a velocity selector apparatus, an electric potential of 350 V is applied, and the equation qV = 1/2 mv^2 is used to calculate the velocity of a proton. The left side of the equation represents the energy gained by a charge q as it moves through a potential difference V, which is the work done by the electric field on the charge. This relationship illustrates that the kinetic energy gained by the proton, represented by 1/2 mv^2, is equal to the work done on it. Understanding this concept is crucial for analyzing the motion of charged particles in electric fields. The discussion emphasizes the connection between electric potential energy and kinetic energy in the context of charged particles.
darksyesider
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In a velocity selector apparatus, i had an electric potential set across two points, and a proton at rest at one end. The potential was 350 V.

My instructor used :

qV = 1/2 mv^2 to find the velocity at the end of the apparatus.

Where did the left hand side of the eqn come from??
 
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darksyesider said:
qV = 1/2 mv^2 to find the velocity at the end of the apparatus.

Where did the left hand side of the eqn come from??
That describes the energy gained by a charge q moving through a potential difference V. (The work done by the field on the charge.)
 
qv=(0.5)mv^2
This expression indicates Kinetic Energy gained by a body with charge q while moving in a region of Electric potential v.
 

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