What is the change in KE of the proton?

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To calculate the change in kinetic energy (KE) of a proton moving through an electric field, the correct formula is KE = qV, where q is the charge of the proton and V is the potential difference. In this scenario, the proton moves 0.1 m in an electric field of 3 V/m, leading to a potential difference of V = Ed = 3 V/m * 0.1 m = 0.3 V. Substituting the values, the change in KE can be calculated as KE = (1.6 x 10^-19 C)(0.3 V), resulting in a change of 4.8 x 10^-20 J. The initial formula presented was incorrect due to misunderstandings of the terms involved. Accurate calculations require using the correct relationships between charge, electric field, and distance.
Dx
A proton moves .1 m along the direction of an electric field of magnitude 3 V/m. What is the change in KE of the proton?

Is this the corrrect formula? KE = Q_pV_accel

Through substition i came up with an swer of 3.2x10^-20J but i don't think its correct?
Can you help me solve for this please?

Thanks!

Dx :wink:
 
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Originally posted by Dx
Is this the corrrect formula? KE = Q_pV_accel

Can you explain what those terms mean?
 


Originally posted by Tom
Can you explain what those terms mean?

Sure Tom,

KE=Kinetic Energy
Q_p=Charge of potential
V_accel=velocity of acceleration

Its not the correct formula huh?
 
Dx,

There is no such thing as a "charge of potential" or "velocity of acceleration". I think you have those concepts muddled up. The kinetic energy K that a charge q acquires while being accelerated through an potential difference* V=Ed is:

K=qV=qE/d

*I can use V=Ed because that is valid for parallel plate capacitors.
 
A proton moves .1 m along the direction of an electric field of magnitude 3 V/m. What is the change in KE of the proton?
proton mass = 1.6e-27 kg
proton charge = 1.6e-19 C
distance = 0.1m
E = 3 V/m
mv2/2 = energy (non-relativistic)
KE=qE/d...
just pick up the calculator.
 

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