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University Physics | 13th Edition | ISBN: 9780321675460 | Authors: Hugh D. Young, Roger A. Freedman
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CALC A vacuum tube diode consists of concentric

Chapter 23, Problem 57P

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CALC A vacuum tube diode consists of concentric cylindrical electrodes, the negative cathode and the positive anode. Because of the accumulation of charge near the cathode, the electric potential between the electrodes is given by V(x) = Cx4/3 where x is the distance from the cathode and C is a constant, characteristic of a particular diode and operating conditions. Assume that the distance between the cathode and anode is 13.0 mm and the potential difference between electrodes is 240 V. (a) Determine the value of C. (b) Obtain a formula for the electric field between the electrodes as a function of x. (c) Determine the force on an electron when the electron is halfway between the electrodes.

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QUESTION:

CALC A vacuum tube diode consists of concentric cylindrical electrodes, the negative cathode and the positive anode. Because of the accumulation of charge near the cathode, the electric potential between the electrodes is given by V(x) = Cx4/3 where x is the distance from the cathode and C is a constant, characteristic of a particular diode and operating conditions. Assume that the distance between the cathode and anode is 13.0 mm and the potential difference between electrodes is 240 V. (a) Determine the value of C. (b) Obtain a formula for the electric field between the electrodes as a function of x. (c) Determine the force on an electron when the electron is halfway between the electrodes.

ANSWER:

Introduction We have to first calculate the constant C. Then we have to find the electric field inside the vacuum tube as a function of x, and then we have to calculate the force on an electron when the electron is in the halfway between the cathode and anode. Step 1 (a) The potential difference at a distance x is given by V (x) = Cx 4/3 3 Now we have V = 240 V when x = 13.0 mm = 13.0 × 10 m. Hence we have (240 V ) = C(13.0 × 10 3 m) 4/3 240 V 4 4/3 C = (13.0×10m)4/3= 7.85 × 10 V /m So the value of the constant is 7.85 × 10 V /m 4/3.

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