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
.
Hello.
The answer is <span>remains the same.
</span>The total amount of energy stays the same because the 1st Law of Thermodynamics states that energy can neither be created nor destroyed, it can only change forms. <span>So the chemical energy is just being converted into heat and light.
</span>
Have a nice day
To solve this problem it is necessary to take into account the concepts of Intensity as a function of Power and the definition of magnetic field.
The intensity depending on the power is defined as

Where
P = Power
r = Radius
Replacing the values that we have,


The definition of intensity tells us that,

Where,
Magnetic field
Permeability constant
c = Speed velocity
Then replacing with our values we have,

Re-arrange to find the magnetic Field B_0

Therefore the amplitude of the magnetic field of this light is
Answer:
Bro where is image Hope you understand me