Two parallel plates are separated by a distance of 45.0mm. The potential differe

Question

Two parallel plates are separated by a distance of 45.0mm. The potential difference across the plates is 25.5 kV.

a.   What is the magnitude of the electric field between the plates?

b.   A beam of electrons leaves the negative plate with an initial velocity of 3.5x107 m/s, what is the change in kinetic energy in electron volts of the electrons when they reach the positive plate?

c.   Using Newtonian mechanics, what is the velocity of the electrons before they strike the positive plate?

mass of electron 9.109x10-31 kg   electronic charge 1.602x10-19 C

Explanation / Answer

A) E d = V

So E = 5.67* 105 N/C

Electric field is directed from positive to negative plate and electron is negatively charged so force on electron is directed from negative to positive plate

So work done = qV = Change in Kinetic energy

put the values we get,

final velocity = 10.1 * 107 m/s

and change in kinetic energy = qV = 4.08 * 10-15 J

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