Answer with Explanation:
a. Option d is true.
a negatively charged plane parallel to the end faces of the cylinder
b. Radius of cylinder, r=0.66m
Magnitude of electric field, E=300 N/C
We have to find the net flux through the closed surface.
Net electric flux,
c.
Net charge,
Where
Where
Answer:
a) 60 V
b) 125 V
c) 125 V
Explanation:
<u>Given</u>
We are given the total electric charge q = 6.75 nC = 6.75x 10^-9 C distributed uniformly over the surface of a metal sphere with a radius of R = 20.0 cm = 0.020 m.
<u>Required </u>
We are asked to calculate the potential at the distances
(a) r = 10.0 cm
(b) r = 20.0 cm
(c) r = 40.0 cm
<u>Solution</u>
(a) Here, the distance r > R so, we can get the potential outside the sphere (r > R) where the potential is given by
V = q/4∈_o (1)
r is the distance where the potential is measured and the term 1/4∈_o equals 9.0 x 10^9 Nm^2/C^2. Now we can plug our values for q and r into equation (1) to get the potential V where r = 0.10 m
V= 1*q/4∈_o*r
=60 V
(b) Here the distance r is the same for the radius R, so we can get the potential inside the sphere (r = R) where the potential is given by
V = 1*q/4∈_o*R (2)
Now we can plug our values for q and R into equation (2) to get the potential V where R = 0.20 m
V = 1*q/4∈_o*R
= 125 V
(c) Inside the sphere the electric field is zero therefore, no work is done on a test charge that moves from any point to any other point inside the sphere. Thus the potential is the same at every point inside the sphere and is equal to the potential on the surface. and it will be the same as in part (b)
V= 125 V
Answer:
The automobile tire rotates 91 revolutions
Explanation:
Given;
angular acceleration of the automobile, α = 2.13 rad/s²
time interval, t = 23.2-s
To calculate the number of revolutions, we apply the first kinematic equation;
the initial angular velocity is zero,
Find how many revolutions that are in 573.2256 Rad
Therefore, the automobile tire rotates 91 revolutions
