a) Consider the path from A to B. along this path, what is the change in the ele

Question

a) Consider the path from A to B. along this path, what is the change in the electric potential?
b) There is a tiny hole in the central plate, so a moving particle can pass through the hole. If an electron moved from A to B along the path shown, what is would be the change in its kinetic energy?
c) What is the minimum kinetic energy the electron must have in order to ensure that it reaches location B? (25 points) As shown in the figure below, three large, thin, uniformly charged plates are arranged so that there are two adjacent regions of uniform electric field. The origin is at the center of the central plate. Location A is (0.4,0, 0) m, and location B is (0.2,0,0) m. The electric field E1 has the value (725,0,0) V/m, and E2 is (425,0,0) V/m. + A E1 E2

Explanation / Answer

Since potential decreases when we move along the direction of field and it increases when we move opposite to it so

Vb -Va = -E1{0-(-0.4)} - E2(0.2-0).....(location of middle plate is 0,0,0)

So

Vb- Va= -290+85= -205 V ..........1)

a.) Change in potential when we move from A to B is -205 volt.

b.)Using conservation of total mechanical energy of electron (charge =-e) we can say that

U+ K =0

So

K = -U= -qV = e(Vb -Va)

K =-205e = -3.28× 10-17 J

c.)

If initial K.E is K than for minimum vaule of K final kinetic energy =0

Hence change in KE K=-K

From the data of part b

We can say

Minimum KE

K=-K=3.28×10-17 J

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