Answer:
8.854 pF
Explanation:
side of plate = 0.1 m ,
d = 1 cm = 0.01 m,
V = 5 kV = 5000 V
V' = 1 kV = 1000 V
Let K be the dielectric constant.
So, V' = V / K
K = V / V' = 5000 / 1000 = 5
C = ε0 A / d = 8.854 x 10^-12 x 0.1 x 0.1 / 0.01 = 8.854 x 10^-12 F
C = 8.854 pF
Answer:
0.67 s
Explanation:
This is a simple harmonic motion (SHM).
The displacement,
, of an SHM is given by

A is the amplitude and
is the angular frequency.
We could use a sine function, in which case we will include a phase angle, to indicate that the oscillation began from a non-equilibrium point. We are using the cosine function for this particular case because the oscillation began from an extreme end, which is one-quarter of a single oscillation, when measured from the equilibrium point. One-quarter of an oscillation corresponds to a phase angle of 90° or
radian.
From trigonometry,
if A and B are complementary.
At
, 


So

At
, 





The period,
, is related to
by

the enclosed may help. earth rotes and you and i are fixed to it. air flows it's own path ...
Answer:
(a) A. Uniform line of charge and B. Uniformly charged sphere
(b) To three digits of precision:
λ = 1.50 * 10^-10 C/m
p = 2.81 * 10^-4 C/m^3
Explanation:
Answer:
The tube should be held vertically, perpendicular to the ground.
Explanation:
As the power lines of ground are equal, so its electrical field is perpendicular to the ground and the equipotential surface is cylindrical. Therefore, if we put the position fluorescent tube parallel to the ground so the both ends of the tube lie on the same equipotential surface and the difference is zero when its potential.
And the ends of the tube must be on separate equipotential surfaces to optimize potential. The surface near the power line has a greater potential value and the surface farther from the line has a lower potential value, so the tube must be placed perpendicular to the floor to maximize the potential difference.