The electron gun in an old TV picture tube accelerates electrons between two par

Question

The electron gun in an old TV picture tube accelerates electrons between two parallel plates 1.3cm apart with a 27kV potential difference between them. The electrons enter through a small hole in the negative plate, accelerate, then exit through a small hole in the positive plate. Assume that the holes are small enough not to affect the electric field or potential.

a)what is the electrical field strength between the plates?

b)With what speed does an electron exit the electron gun if its entry speed is close to zero?

Explanation / Answer

electric field and Potential V are related by E = V/d

where d is the distance between the plates

so

E = 27000/0.013

E = 2.076 *10^6 V/m or N/C-------<<<<<<<< answer to part A

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apply Kinetic energy KE = Potential eenrgy PE

i.e 0.5 mv^2 = eV

v^2 = 2eV/m

where e is the charge   = 1.6 e-19 C

and m is mass of eelctron = 9.11 e-31 kgs

so

v^2 = (2 * 1.6 e-19 * 27000/9.11e-31)

v^2 = 94.84 *10^14

v = 9.738 *10^7 m/s --------- <<<<<<< answer to part

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