The given question is incomplete. The complete question is as follows.
A parallel-plate capacitor has capacitance
= 8.50 pF when there is air between the plates. The separation between the plates is 1.00 mm.
What is the maximum magnitude of charge that can be placed on each plate if the electric field in the region between the plates is not to exceed
V/m?
Explanation:
It is known that relation between electric field and the voltage is as follows.
V = Ed
Now,
Q = CV
or, Q = 
Therefore, substitute the values into the above formula as follows.
Q = 
=
= 
Hence, we can conclude that the maximum magnitude of charge that can be placed on each given plate is
.
The magnitude of the magnetic moment due to the electron's motion is
.
<h3>
What is magnetic moment?</h3>
The magnetic pull and direction of a magnet or other object that produces a magnetic field are referred to as the magnetic moment in electromagnetism. Things that have magnetic moments include electromagnets, permanent magnets, various compounds, elementary particles like electrons, and a number of celestial objects (such as many planets, some moons, stars, etc).
The term "magnetic moment" really refers to the magnetic dipole moment of a system, which is the portion of the magnetic moment that can be represented by an equivalent magnetic dipole or a pair of magnetic north and south poles that are only very slightly apart. The magnetic dipole component is adequate for sufficiently small magnets or over sufficiently large distances.
Calculations:
radius= 
velocity=
Working formula, M=N/A


=


=
M=
=
To learn more about magnetic moment ,visit:
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Answer:
The minimum work per unit heat transfer will be 0.15.
Explanation:
We know the for a heat pump the coefficient of performance (
) is given by

where,
is the magnitude of heat transfer between cyclic device and high-temperature medium at temperature
and
is the required input and is given by
,
being magnitude of heat transfer between cyclic device and low-temperature
. Therefore, from above equation we can write,

Given,
and
. So, the minimum work per unit heat transfer is given by

Answer:
D. 5m
Explanation:
fλ = c, where f is frequency, λ is wavelength and c is speed.
6λ=30
λ=30/6=5