Question

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

Answer 1

v = 138,564 m/s

Step-by-step explanation:

• 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:
• 
  `\Delta U = \Delta V * q_(p) = E*d*q_(p) = 50,000N/C*0.002m*1.6e-19C`
• So, we can write the following equality:

`\Delta U = \Delta K (1)`

where ΔK = 1/2*m*vp² (2)

• Replacing by the values in ΔU, and solving for v, we get:
• 
  `v_(p) =\sqrt{(2*E*d*q_(p))/(m)} = \sqrt{(3.2e10)/(1.67) } m/s = 138,564 m/s`