1.) Consider three charges q 1 = 5.7 nC, q 2 = 7.9 nC, and q 3 = - 3.1 nC, arran
ID: 2280267 • Letter: 1
Question
1.) Consider three charges q1 = 5.7 nC, q2 = 7.9 nC, and q3 = -3.1 nC, arranged in a triangle as shown below.
(a) What is the electric force acting on the charge at the origin?
____ N, ______
Consider three charges q1 = 5.7 nC, q2 = 7.9 nC, and q3 = -3.1 nC, arranged in a triangle as shown below. What is the electric force acting on the charge at the origin? What is the net electric field at the position of the charge at the origin? The classic Millikan oil drop experiment was the first to obtain an accurate measurement of the charge on an electron. In it, oil drops were suspended against gravity by a vertical electric field. Assume the oil drop to be 2.50 mu m in radius and have a density of 887 kg/m3. Find the weight of the drop. If the drop has a single excess electron, find the magnitude of the electric field strength needed to balance its weight. Newer automobiles have filters that remove fine particles from exhaust gases. This is done by charging the particles and separating them with a strong electric field. Consider a positively charged particle +3.1 mu C that enters an electric field with strength 6 ? 106 N/C. The particle is traveling at 17 m/s and has a mass of 10?9 g. What is the acceleration of the particle?Explanation / Answer
1.a) -kq1q2/r2^2 i - kq1q3/r3^2 j ,
= -k*5.7*7.9/(0.3)^2 i - k*5.7*3.1/(0.1)^2 j
=-4.503*10^-6 i - 15.903*10^-6 j
magnitude=16.53*10^-6 N. and angle= 74.19 degrees.
b) Net electric field = F/q1= 2.9810^3 N/C. ang angle=74.19 degrees.
2.a) weight=density*volume*g
volume=4/3*pi*r^3
weight=5.689*10^-13 N.
b) weight will be balanced by electric force.
W=qE
E=W/q=5.689*10^-13/(1.6*10^-19) =3.55*10^6 N/C.
3)ma=qE
a=qE/m=3.1*10^-6 * 6*10^6/10^-12 =18.6 * 10^12 m/s^2 downwards.
Related Questions
Navigate
Integrity-first tutoring: explanations and feedback only — we do not complete graded work. Learn more.