the electric field strength is 50,000 n/c inside a parallel-plate capcitor with a 2.0 mm spacing. a proton is released from rest at the positive plate. what is the proton's speed when it reache the negative plate
The potential difference between both plates of the capacitor, is just the work done on the proton by the electric field, per unit charge.
This work must be equal to the change in the kinetic energy of the proton, going from rest at the positive plate to some speed at the negative plate.
The change in the electrostatic potential energy can be expressed as follows, taking into account that the electric field is the electrostatic force per unit charge:
[tex]\Delta U = \Delta V * q_{p} = E*d*q_{p} = 50,000N/C*0.002m*1.6e-19C[/tex]
So, we can write the following equality:
[tex]\Delta U = \Delta K (1)[/tex]
where ΔK = 1/2*m*vp² (2)
Replacing by the values in ΔU, and solving for v, we get:
[tex]v_{p} =\sqrt{\frac{2*E*d*q_{p}}{m}} = \sqrt{\frac{3.2e10}{1.67} } m/s = 138,564 m/s[/tex]
