A) What must the charge (sign and magnitude) of a particle of mass 1.49 g be for
ID: 1706465 • Letter: A
Question
A) What must the charge (sign and magnitude) of a particle of mass 1.49 g be for it to remain stationary when placed in a downward-directed electric field of magnitude 600 N/C?Use 9.81 m/s^2 for the magnitude of the acceleration due to gravity.
B) What is the magnitude of an electric field in which the electric force on a proton is equal in magnitude to its weight?
Use 1.67×10-27 kg for the mass of a proton, 1.60×10-19 C for the magnitude of the charge on an electron, and 9.81 m/s^2 for the magnitude of the acceleration due to gravity.
Explanation / Answer
a)
mass of the particle m =1.49 g =1.49*10^-3 kg
the electric field E =600 (-j)N/C
(since electric field in downward direction)
magnitude of the electric field IEI =600 N/C
acceleration due to gravity g =9.81 m/s^2
the relation between electric field and electro static force is
F =qE
charge q =F/E ...... (1)
but , from Newton's second law of motion,
force F =mg ....... (2)
therefore, magnitude of charge q =mg/E ....... (3)
q =24.36*10^-6 C
q =24.36 C (positive)
.................................................
b)
from Newton's second law of motion,
force F =mg ....... (4)
the relation between electric field and electro static force is
F =qE ......... (5)
compare eq (4) and (5), we get
qE =mg
electric field E =mg/q ....... (6)
where, mass of proton m =1.67*10^-27 kg
charge of proton q =1.6*10^-19 C
and g =9.81 m/s^2
magnitude of the electric field E =mg/q
=10.23*10^-8 N/C
Related Questions
drjack9650@gmail.com
Navigate
Integrity-first tutoring: explanations and feedback only — we do not complete graded work. Learn more.