1. An ideal parallel plate capacitor with a plate separation of d = 5.00 x i0 m
ID: 1462900 • Letter: 1
Question
1. An ideal parallel plate capacitor with a plate separation of d = 5.00 x i0 m and a capacitance of C = 4.00 x 106 F is charged so that the total charge on the upper plate is Q = +2.00 x i0 C, as shown in the figure. An electron moving to the right with a speed of y = 2.50 x 106 rn/s enters the region between the plates. What magnitude and direction of magnetic field should you apply in the region of the capacitor such that the electron is undeflected? A) 2.00 x 10^-8 T out of the page B) 2.00 x 10^-8 T into the page C) 4.00 x 10^08 T out of the page D) 4.00 x 10^-8 T into the page E) None of the aboveExplanation / Answer
Given
Separation of plates d = 5 x 10-3m
Charge on plates Q = 2 x 10-9 C
Capacitance of the plate C = 4 x 10-6 F
Velocity of the electron v = 2.50 x 106 m
Solution
Electric field between the plates E = potential difference V / Seperaion d
Potential difference v = Charge Q/Capacitance C
E = Q/Cd
= 2 x 10-9/4 x 10-6x5 x 10-3
= 0.1 N/m
Deflecting force action on electron by electric field F = Ee
The magnetic force must be equal to this force
Ee = Bev
E = Bv
B = E/v
= 0.1/2.50 x 106
B= 0.04 x 10-6 T = 4 x 10-8 T
Using Fleming left hand rule the magnetic field should be directed out of the paper to keep the electron from deflecting towards positively charged upper plate