Capacitance

[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 dont think its correct?
Can you help me solve for this please?

Thanks!

Dx [;)]

[Tom Mattson]

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


Can you explain what those terms mean?

[Dx]

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?

[Tom Mattson]

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.

[schwarzchildradius]

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.