In the electron gun pictured, a potential difference of 12.9 kV is found to brin

Question

In the electron gun pictured, a potential difference of 12.9 kV is found to bring electrons from rest at the cathode to 6.7 107 m/s at the anode. If the electrons reach the anode with a speed of 2.9 107 m/s, what is the potential difference between the cathode and the anode? kV

Electron Plates for gun horizontal deflection (B) Plates for vertical deflection Heated filament (source of electrons) Cathode Uniform E field Electron beam seen from side Cathode Conductive coating Anode Side view Fluorescent screen

Explanation / Answer

Voltage accelerates the electrons, hence by using work energy equation we can find out the Potential difference.

Work Done by Electrical Force = Change in Kinteic Energy

q V = 0.5 x m (Vf2 - Vi2)

Vi = 0

1.6 x 10-19 x V = 0.5 x 9.1 x 10-31 x (2.9 x 107)2

V = 2391.59 V

= 2.39 kV

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