A uniform electric field of magnitude 370 N/C pointing in the positive x-direction acts on an electron, which is initially at rest. The electron has moved 2.70 cm.

Question

(a) What is the work done by the field on the electron?
J?

(b) What is the change in potential energy associated with the electron?
J?

(c) What is the velocity of the electron?
magnitude? m/s
direction



2.
Oppositely charged parallel plates are separated by 5.26 mm. A potential difference of 600 V exists between the plates.
(a) What is the magnitude of the electric field between the plates?
N/C?

(b) What is the magnitude of the force on an electron between the plates?
N?

(c) How much work must be done on the electron to move it to the negative plate if it is initially positioned 2.94 mm from the positive plate?
J?

Answer 1

1 A) W = e*E*X
X must be in meters to get the Work in Joules. e is the elctron chrge
B) -W
C) (1/2) m v^2 = W
Solve for velocity v

2. (a) E = V/d
d must be in meters
(b) e*E
(c) It is 2.32 cm = 0.0232 m from the positive plate. Call that distance X'

Work = e E x'

Answer 2

How do you determine the electric charge? What does X represent?

Answer 3

Actually I know them now but what is M for C) on the third question?

Answer 4

for the first question i mean

Answer 5

M is the mass of the electron.
9.108*10^-31 kg