Board work A charge moving at an unknown speed enters a uniform electric field c

Question

Board work A charge moving at an unknown speed enters a uniform electric field created between two charged plates. Plate A is held at 0 V, and Plate Bis held at 12 The plates are separated by 4 cm, and the charge stops 2 cm from Plate B. (a) Draw the direction of the electric field created Plate B Plate A by these charged plates (b) Is the charge a proton or an electron? (c) Calculate the magnitude of the electric field between the plates. (d) Calculate the charge's speed when it entered the electric field. (e) The two plates creating this electric field make up a 4 cm 50 HF capacitor. How many electrons are on the negatively charged plate? 12 V 0 V Answer: a) Left, (b) Proton; (c) E 300 VIm; (d) v 3.4 x 104 m/s; (e) 3.8 x 1015 electrons

Explanation / Answer

a.

electric field lines move from + terminal to - ve terminal

i.e from high potential to low potential

as B is at high potential , Field lines wud travel from Plate B to Plate A .

0V !<--------------------<------------! 12V          
!<--------------------<------------!
!<--------------------<------------!          
!<--------------------<------------!
!<--------------------<------------!   
!<--------------------<------------!

part B :

Since partilce in moving in to the field , this must be a +ve partlce as this is attarected in the region
of electric field

part C:

Electric FIeld E , Electric potential V and distance dare related by

E = V/d

E = (12 -0)/(0.04)

E = 300 V/m

part D :

use gravitational force = electrical force

ma = E q

a = 300* 1.6*10^-19/(1.67*10^-27)

a = 2.87 *10^10 m/s^2

V^2 = 2* 2.87 *10^10 * (0.04 -0.02)

V = 3.38 *10^4 m/s ~ 3.4 *10^4 m/s

part E:

charge Q = CV

Q = 50*10^-6 * 12

Q = 600*10^-6 C

as Q = n e

n = Q/e

n = 600*10^-6/(1.6*10^-19)

n = 3.8*10^15 electrons

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