Question

Suppose that in an electron gun, electrons are emitted from a hot filament at point f and are accelerated by an electric field to an opening at o, as shown. The charge on an electron is (-e) where e = 1.6 times 10^-19 C. (a) If the voltage (same as electric potential) at point f is - 280 Volts, what is the potential energy of an electron at f? (b) The electrons come off the filament with very low velocity, hence with very little kinetic energy. So what is the total energy of an electron at f? (c) After the electrons exit the gun at o the voltage is zero. Thus, what is the potential energy of an electron at o? (d) Given that the mass of an electron is m = 9.11 times 10^-31 kg, how fast must an electron be going when it exits the gun?

Answer 1

Given,
The value of charge = charge of electron = 1.6X 10^(-19) C
The electric potential at the point f = -280 V

a) as we know, The potential energy of a charge "q" subjected to an electric potential "V" is given by,
   P.E. = q*V
   = (-1.6X10^(-19)) * (-280) =

4.48 10-17j

b) the total energy at point f = Kinetic energy + Potential Energy
   = (1/2)(m)(V^2) +
4.48 10-17j

=(1/2)(m)(0^2) +
4.48 10-17j

   =
4.48 10-17j

c) The potential energy of a charge "q" subjected to an electric potential "V" is given by,
   P.E. = q*V . Here at the point o, the voltage is 0 V
   = (-1.6X10^(-19)) * (0) = 0 J

d) By the principle of conservation of energy, the total energy at point f is same as the energy at point o.

or,
  KEi + PEi = KEf + PEf
0 +
4.48 10-17j
= KEf + 0 J
Now, KEf = $\(1/2)*m_{e}*V_{e}^{2}$

or , =


or,


Please let me know if any questions or queries.