(Parte 9 de 11)

qx mvEy

Substitute the non-particle specific q mEy

(a) Substitute for the mass and charge of an electron and evaluate

Ey:

kN/C3.21

(b) Substitute for the mass and charge of a proton and evaluate Ey: ()

The Electric Field 1: Discrete Charge Distributions 39

5 •• Picture the Problem We can use constant-acceleration equations to express the x and y coordinates of the electron in terms of the parameter t and Newton’s 2nd law to express the constant acceleration in terms of the electric field. Eliminating the parameter will yield an equation for y as a function of x, q, and m. We can decide whether the electron will strike the upper plate by finding the maximum value of its y coordinate. Should we find that it does not strike the upper plate, we can determine where it strikes the lower plate by setting y(x) = 0.

Express the x and y coordinates of the

Apply Newton’s 2nd law to relate the acceleration of the electron to the net force acting on it: eye y meEm Fa==net,

Substitute in the y-coordinate equation

tvy e y−=θ

Eliminate the parameter t between the

tan x vm xxyeyθθ−= (1)

To find ymax, set dy/dx = 0 for extrema:

extremafor0

−= x' vm eEdxdy e y θθ

Solve for x′ to obtain:

ye eE vmx' 2

Substitute x′ in y(x) and simplify to

obtain ymax: ye eE vmy 2

Substitute numerical values and evaluate ymax:

and, because the plates are separated by 2 cm, the electron does not strike the upper plate.

Chapter 21 40

To determine where the electron will strike the lower plate, set y = 0 in equation (1) and solve for x to obtain:

ye eE

Substitute numerical values and evaluate x:

Remarks: x′ is an extremum, i.e., either a maximum or a minimum. To show that it is a maximum we need to show that d2y/dx2, evaluated at x′, is negative. A simple alternative is to use your graphing calculator to show that the graph of y(x) is a maximum at x′. Yet another alternative is to recognize that, because equation (1) is quadratic and the coefficient of x2 is negative, its graph is a parabola that opens downward.

56 •• Picture the Problem The trajectory of the electron while it is in the electric field is parabolic (its acceleration is downward and constant) and its trajectory, once it is out of the electric field is, if we ignore the small gravitational force acting on it, linear. We can use constant-acceleration equations and Newton’s 2nd law to express the electron’s x and y coordinates parametrically and then eliminate the parameter t to express y(x). We can find the angle with the horizontal at which the electron leaves the electric field from the x and y components of its velocity and its total vertical deflection by summing its deflections over the first 4 cm and the final 12 cm of its flight.

(a) Using a constant-acceleration equation, express the x and y coordinates of the electron as functions of time:

()tvtx0=(1)
and ()221tatyy=

Because v0,y = 0:

Using Newton’s 2nd law, relate the acceleration of the electron to the electric field:

e y ey meEm

The Electric Field 1: Discrete Charge Distributions 41

Substitute to obtain:

y−=(2)

ty e

Eliminate the parameter t between equations (1) and (2) to obtain: () 2 eE x

Substitute numerical values and evaluate y(4 cm):

(b) Express the horizontal and vertical components of the electron’s speed as it leaves the electric field:

θcos0vvx= and θsin0vvy=

Divide the second of these equations by the first to obtain: 0

1 tantan v v yx

Using a constant-acceleration equation, express vy as a function of the electron’s acceleration and its

time in the electric field:

tavvyyy+=,0 or, because v0,y = 0

0net, vxm eE t tav eye xeE

vm xeE yey 2 tantan 1 2

Substitute numerical values and evaluate θ :

(c) Express the total vertical

Relate the horizontal and vertical distances traveled to the screen to the horizontal and vertical components of its velocity:

(Parte 9 de 11)

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