A proton is released from rest within a horizontal electric field E_h and procee

Question

A proton is released from rest within a horizontal electric field E_h and proceeds to accelerate to the right.

A) If it accelerates at 1e^12 m/s^2, what is the magnitude and direction of E_h?

B) The length of the region with E_h is 20 mm. What is its final horizontal speed when it exits that region

C) The proton then enters a vertical electric field E_v, where the electric field strength is 3e^4 N/C upwards. If the proton spends 2.5e^-7 s in the new field, how far (t) does it travel to the right?

D) how much is deflected (delta y) from its original path and in what direction (up or down)?

Explanation / Answer

part A :

use the balancing condition as Fgrav = F eelc

ma = Eq

where m is mass of proton = 1.67 e-27 kgs

q is charge = 1.6 e =19 C

a is accleration = 1e 12 m/s^2

so

Eletric field E = ma/q

E = 1.67 e -27 * 1 e 12 /(1.6 e -19)

E = 1.04 *10^ 4 N/C

----------------------------------------

use V^2 = 2aS

V^2 = 2* 1.e 12 * 20 e -3

V = 2 e 5 m/s

-------------------------------------------------------

accleration a = Eq/m

a = 3e4 * 1.6e -19/1.67 e-27

a = 2.87 e 12 m/s^2

so

Vertical Heght H = 0.5 at^2

H = 0.5* 2.87 e 12 * 2.5 e -7 * 2.5 e -7

H = 8.9 cm

-------------------------------------------------
change of accleration a = 2.87 -1 = 1.87 e 12 , m/s^2

so

H defelc = 0.5* 1..87e 12 * 2.5 e -7 * 2.5 e -7

Hdefec = 5.8 cm

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