Solution Found!
Figure 19?41 shows an electron entering a parallel?plate
Chapter 19, Problem 101GP(choose chapter or problem)
Figure 19–41 shows an electron entering a parallel-plate capacitor with a speed of
\(5.45 \times 10^{6} \mathrm{~m} / \mathrm{s}\). The electric field of the
capacitor has deflected the electron downward by a distanceof 0.618 cm at the point where the electron exits the capacitor. Find
(a) the magnitude of the electric field in the capacitor and
(b) the speed of the electron when it exits the capacitor.
Equation Transcription:
Text Transcription:
5.45 X 106 m/s
Questions & Answers
QUESTION:
Figure 19–41 shows an electron entering a parallel-plate capacitor with a speed of
\(5.45 \times 10^{6} \mathrm{~m} / \mathrm{s}\). The electric field of the
capacitor has deflected the electron downward by a distanceof 0.618 cm at the point where the electron exits the capacitor. Find
(a) the magnitude of the electric field in the capacitor and
(b) the speed of the electron when it exits the capacitor.
Equation Transcription:
Text Transcription:
5.45 X 106 m/s
ANSWER:Step 1 of 5
(a)
Here we need to calculate the magnitude of the electric field in the parallel plate capacitor.
The electric field of the capacitor has deflected an electron which is moving with a speed of downwards by a distance of .The length of the capacitor is given to be . The magnitude of charge and mass of the electron are taken to be and respectively.