Need help on A, C, D, and E The electric force experienced by a -5.6 micro coulo

Question

Need help on A, C, D, and E

The electric force experienced by a -5.6 micro coulomb charge at some point P has a magnitude of 18.8 N and points due North.

i safari File Edit View History Bookmarks Window Help ) 25% I- Tue 11:58 PM a Pandora SoundCloud Yahoo Amazon Google YouTube Facebook Netflix csun portal Moodle Canvas Mastering Physics Expert TA Assignment Status Click here for detailed view 20% Part (a) what is the magnitude of the electric field, in newtons per coulomb, at P? Grade Summary 0% 100% Potcntial Problem Status Submissions Attempts remaining: 16 (0% per attempt) sinO tan() | ( cosO cotanasin acos0 atanacotan sinh cosh0 cotanh0 Degrees. Radians 1 Completed 2 Completed detailed view 2 3 0% 0% 0% 0% 4 Completed 5 Completed END Submit I give up Hints: 0 for a 0% deduction. Hints remaining: Feedback: deduction per feedback. Submission History lints Totals E--3.21103 0% 0% 0% | 0% 0% | 0% 0% | 0% E--321000 E--3.36 0% 0% Note: Feedback not accessed. E =-3.36 E- 3.36 E3.36 0% 0% 0% 0% Note: Feedback not accessed. Note: Feedback not accessed. 106 0% 0% E- -3.36E+6 Totals g 20% Part (b) what is the direction of the electric field at point P? given charged particle? experience? has the same strength? 0% 0% 0% 0% 20% Part (c) what is the magnitude ofthe force, in newtons, that an electron would experience if it were placed at point P instead of the 20% Part (d) If a proton were placed at point P instead of the electron, what is the magnitude of the force, in newtons, that it would 20% Part (e) what is the ratio of the magnitude of the acceleration of an electron to that of a proton, when the electric field acting on each ll con:ent&2017 Expert T

Explanation / Answer

part a )

F = qE

E = F/q

F = 18.8 N

q = -5.6 x 10^-6 C

|E| = 3.357 x 10^6 N/C

part c )

they are in same electric field

F = qE

F1 = q1*E

F2 = q2*E

F1/F2 = q1/q2

q2 = electron charge = 1.6 x 10^-19 C

F1 = 18.8 N

q1 = 5.6 x 10^-6 C

F2 = F1*q2/q1

F2 = 5.37 x 10^-13 N

part d )

it will be same in magnitude of part c because proton has same magnitude of charge of electron

part e )

F = ma = qE

for elecrton

Fe = me*ae

for proton

Fp = mp*ap

Fe/Fp = me*ae/mp*ap

mp/me = ae/ap

mp = mass of proton = 1.67 x 10^-27 kg

me = mass of electron = 9.11 x 10^-31 kg

ae/ap = 1833

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