i have the correct answers and equations, i just am not ending up with those ans
ID: 1699108 • Letter: I
Question
i have the correct answers and equations, i just am not ending up with those answers.....Two parallel plates have dimension of 25cm x 36cm, adn they each have voltage of 75V applied across them
a)if the plate separation is 4mm, what is the electric field between the plates?
***E=18,750V/m
b)what is the surface charge density on each of teh plates?
***1.66(10^7)C/m^2
c)what is total charge on each plate?
***Q=.015
d)if a proton is released from the positive plate, what speed will it acquire when it hits the negative plate.
***v=2(10^5)m/s
Explanation / Answer
Given
area of plates A = 0.25*0.36 m^2 = 0.09 m^2
potential across the plates V = 75 V
a) the distance between the plates d = 4*10^-3 m
Electric field between the plates E = V / d
= 75 / 4*10^ -3
= 18750 V / m
b) surface charge density on each plate
= E _o
= 18750 * 8.85*10^ -12 = 1.659*10^ -7 C / m^2
c) charge on each plate
Q = A
= 1.659 *1^ -7 * 0.09 = 0.149 *10^ - 7 C
d) 1 / 2 m v^2 = V q
1 / 2 ( 1.6*10^ -27 ) v^2 = 75 *1.6*10^ -19
v^2 = 150*10^ 8
v = 12.24*10^ 4
= 1.224 *10^ 5 m/s
Related Questions
Navigate
Integrity-first tutoring: explanations and feedback only — we do not complete graded work. Learn more.