The electric force experienced by a -4.4 micro Coulomb charge at some point P ha
ID: 1656145 • Letter: T
Question
The electric force experienced by a -4.4 micro Coulomb charge at some point P has a magnitude of 19.8 N and points due north. A). What is the magnitude of the electric field, in newtons per coulomb, at P? B). What is the direction of the electric field at point P? C). What is the magnitude of the force, in newtons, that an electron would experience if it were placed at point P instead of the given charged particle? D). If a proton were placed in point P instead of the electron, what is the magnitude of the force, in newtons, that it would experience? 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 has the same strength?
Explanation / Answer
A] |E| = F/q = 19.8/(-4.4u) = 4.5 x 106 N/C.
B] Since the charge is negative and it experiences a force in North direction, therefore the electric field is in the south direction.
C] |F| = eE = 1.6 x 10-19 x 4.5 x 106 = 7.2 x 10-13 N.
D] Since both the electron and proton have same magnitude of charge: |q| = 1.6 x 10-19 C therefore the force on the proton will also be: |F| = 7.2 x 10-13 N.
E] F = ma
so since |F| is same for both electron and proton therefore their ratio of accelerations should be equal to the ratio of their masses.
=> ae/ap = mp/me = 1758.