please show all steps, and justify the reason behind all qeuations used. 3. A pa

Question

please show all steps, and justify the reason behind all qeuations used.

3. A parallel-plate capacitor is formed of two 10 cm x 10 cm plates spaced 1.0 cm apart. The plates are charged to +/- 1.0 nC. An electron is shot through a very small hole halfway up the positive plate. a. What is the slowest speed the electron can have as it enters the capacitor if it is to reach the negative plate? b. What is the electric potential the electron experiences inside the capacitor when it is a distance of 0.3 cm away from the positive plate? c. Now, a second vertical electric field is applied upwards with E = 2300 N/C. Where does the electron land? (Either the x-position on the top/bottom of the capacitor or the y-position on the negative plate.)

Explanation / Answer

a) there will be ELectric field inside the capacitor and potential difference also.

Potential difference will do work to stop electron.

if potential diff inside the capacitor is V then work done by P.D. on e will be = qV

Capacitance of parallel plate capacitor C = e0*A / d

C = 8.854 x 10^-12 x (0.10 x 0.10) / 0.01

C = 8.854 x 106-12 F

now using

Q = CV

1 x 10^-9 = 8.854 x 10^-12 x V

V = 112.94 Volt

INitial K.E. of electron = work done on electron

mv^2 /2 = qV

9.109 x 10^-31 x v^2 /2 = 1.6 x 10^-19 x 112.94

v = 6.30 x 10^6 m/s

b) electric field throughout the capacitor is same

so E = V / d

E = 112.94 V / 0.01 = 11294 V/m

Potential difference between the +ve plate and charge is E.d = 11294* 0.3*10^-2 = 33.83 Volt

from -ve plate or potential of e = 112.94 - 33.88 =79.06 Volt

c) `electric field in horizontal diretion is Ex = 11294 N/C

Fx = q*Ex

a_x = Fx / m = q*Ex / m

= 1.984 x 10^15 m/s2 in opposite direction of initial velocity.

Initia velocity in horizontal direction ux = 6.30 x 10^6 m/s

final horizontal velocity = 0

using v = u + at

t = u / a = 6.30 x 10^6 / 1.984 x 10^15 = 3.176 x 10^-9 sec

electron will travel for same time in vertical diection with Ey = 2300 N/C in upwards direction.

ay = qEy / m = 4.04 x 10^14 m/s2 downwards

dy = 0 + ay*t^2 /2

= 4.04 x 10^14 x (3.176 x 10^-9)^2 /2 = 2.038 x 10^-3 m

so y position is 2.038 mm below the x-axis ( on - Y axis )

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