Answer:
How far will the electron travel beforehitting a plate is 248.125mm
Explanation:
Applying Gauss' law:
Electric Field E = Charge density/epsilon nought
Where charge density=1.0 x 10^-6C/m2 & epsilon nought= 8.85× 10^-12
Therefore E = 1.0 x 10^-6/8.85× 10^-12
E= 1.13×10^5N/C
Force on electron F=qE
Where q=charge of electron=1.6×10^-19C
Therefore F=1.6×10^-19×1.13×10^5
F=1.808×10^-14N
Acceleration on electron a = Force/Mass
Where Mass of electron = 9.10938356 × 10^-31
Therefore a= 1.808×10^-14 /9.11 × 10-31
a= 1.985×10^16m/s^2
Time spent between plate = Distance/Speed
From the question: Distance=1cm=0.01m and speed = 2×10^6m/s^2
Therefore Time = 0.01/2×10^6
Time =5×10^-9s
How far the electron would travel S =ut+ at^2/2 where u=0
S= 1.985×10^16×(5×10^-9)^2/2
S=24.8125×10^-2m
S=248.125mm
Answer:
the active region is bound by cutoff region and saturation or power dissipation region.
Explanation:
Answer:
<em>The magnitude of the force is 10 N</em>
Explanation:
<u>Coulomb's Law</u>
The electrostatic force between two charged objects is directly proportional to the product of their charges and inversely proportional to the square of the distance between the two objects.
Written as a formula:

Where:

q1, q2 = the objects' charge
d= The distance between the objects
We have two identical charges of q1=q2=1 c separated by d=30000 m, thus the magnitude of the force is:


F = 10 N
The magnitude of the force is 10 N
Answer:
The linear velocity of the object is 8.71 m/s.
Explanation:
Given;
mass of the object, m = 1 kg
radius of the circle, r = 3.3 meters
centripetal force, F = 23 N
Centripetal force is given by;

where;
v is the linear velocity of the object

Therefore, the linear velocity of the object is 8.71 m/s.