An investigation is performed using a apparatus. Electrons are accelerated from
ID: 2008445 • Letter: A
Question
An investigation is performed using a apparatus. Electrons are accelerated from rest from the cathode by an electric potential difference. The cathode and anode are 2.00 x 10 ^-2 cm apart. The electrons reach a speed of 2.10 x 10^7 m/s as they pass through the hole in the anode.
Electrons then pass undeflected through a region in which there is both an electric and a magnetic field. The electron field is produced between two plates ( plate 1 and plate 2) which are 3.00cm apart. Plate 1 is postively charged and plate 2 is negatively charged. The electric potential difference between the plates is 12.0 V. Two current- carrying coils produce a magnetic field that is perendicular to the electric field.
1) What is the electric potential difference between the cathode and the anode in Volts?
2) The electric field between plate 1 and plate 2 is
a) 4.00 x 10^2 toward plate 1
b) 4.00 x 10^2 toward plate 2
c) 4.00 x 10^4 toward plate 1
d) 4.00 x 10^4 toward plate 2
Explanation / Answer
1) Given speed of electron v = 2*107 m/s mass of electron is m = 9.1*10-31 kg potential energy of electron between the plates ( 1/2) m v 2 = V q 0.5 ) ( 9.1*10-31 ) (2*107 ) 2 = V ( 1.6 *10 -19 C ) potential difference between the plates V = 1137.5 V 2) potential difference between the plates is 12 V seperation between the plates d = 3 cm electric field between the plates E = V / d = 12 / 3*10-2 = 4*102 N /C toward plate 2 answer is part B 2) potential difference between the plates is 12 V seperation between the plates d = 3 cm electric field between the plates E = V / d = 12 / 3*10-2 = 4*102 N /C toward plate 2 answer is part BRelated Questions
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