## Saturday, October 29, 2022

### Electric Field Energy: UNIZOR.COM - Physics4Teens - Waves - Energy of Waves

Notes to a video lecture on http://www.unizor.com

Electric Field Energy

Let's calculate the potential energy density PE of an electric field inside a capacitor as a function of the field's intensity E, which we assume to be uniform between the plates of a capacitor.

While the methodology will depend on the fact that this electric field is between the plates of a capacitor, the final formula will depend only on the field's intensity and, as a field's local characteristic, will be the same, no matter what is the outside source of an electric field, whether it's plates of a capacitor or a few point charges, or the result of electromagnetic oscillations far from the source of these oscillations.

Consider a battery used to charge a capacitor from initial no electric charge between plates with the difference of potential between the plates being equal to zero to some charge Qmax with corresponding difference of potential between the plates (voltage) being equal Vmax. This process of charging requires some work performed by the battery.
The total energy accumulated inside a capacitor as a result of charging a capacitor should be equal to this amount of work.

This work performed by a battery to charge a capacitor from 0 voltage to Vmax can be considered a function of accumulated charge W=W(Q) and it grows from zero to some maximum Wmax - the energy accumulated by a capacitor as it accumulated a charge of Qmax.
During the charging process the voltage between the plates of a capacitor can also be considered as a function of accumulated charge Q, that is V=V(Q), and it grows from zero to Vmax.

Recall that the difference in electric potential between two points of an electric field (voltage) is the amount of work needed to transfer a unit of electric charge (1 coulomb) from one point to another.
Therefore, to transfer an additional infinitesimal amount of electric charge dQ from one plate of a capacitor to another, when there is already transferred amount of electric charge Q that creates a voltage V(Q) between the plates, the battery has to spend an additional amount of work
dW(Q) = V(Q)·dQ

Further recall that the electric charge on each plate of a capacitor (positive on one plate and negative on another) Q and the voltage between the plates V are proportional to each other with a capacitor's capacitance C being a factor:
Q = V·C or V = Q/C

Therefore, the above expression for an additional amount of work can be written as
dW(Q) = [Q/C]·dQ

To calculate a total amount of energy Wmax needed to charge a capacitor from zero to Qmax, we have to integrate dW from 0 to Qmax:
Wmax = [0,Qmax][Q/C]·dQ =
= ½·Q²max/C

Of course, the amount of charge Qmax can be any from zero to some practical maximum, so we can drop an index max, and the work spent by a battery will be
W = ½·Q²/C = ½V²·C

Let's approach the same problem from a different viewpoint that involves the intensity E of an electric field between the plates of a capacitor charged with Q amount of electricity to a voltage V between its plates.

The capacitance C of a capacitor was discussed in a lecture "Electromagnetism" - "Electric Field" - "Capacitors" of this course and, as was shown there, depends on the area of each plate A, distance between plates d and electric permittivity of a medium between the plates ε:
C = ε·A/d
with the vacuum having the permittivity ε0 and any other medium having it as
ε = εr·ε0
where εr is relative permittivity of a medium.

Using this expression for a capacitance C, the total work to charge a capacitor with Q amount of electricity to a voltage level V can be written as
W = ½·Q²·d/(ε·A) = ½V²·ε·A/d

The definition of the electric field's intensity E is the force with which a field acts on a unit charge, while voltage V between plates is an amount of work needed to move a unit of charge from one plate to another.

Therefore, using the principal
Work = Force ⨯ Distance,
we can write
V = E·d

Substituting this into a formula for the total work yields
W = ½E²·d²·ε·A/d = ½ε·E²·(A·d)

Notice that A·d is a volume of the space between the plates of a capacitor, occupied by an electric field.
Therefore, the expression W/(A·d) characterizes the density PE of a potential energy of an electric field between the plates of a capacitor.

Hence, the potential energy density of an electric field is a local characteristic that depends on the field's intensity and the permittivity of a medium where the field propagates
PE = ½ε·E²

Final comment is related to the fact that we used a simple kind of an electric field to derive the formula above - the static uniform finite field between two plates of a capacitor. The formula contains only the local property of the field - its intensity at any point E.
Even if the field is of a more complex kind (variable, non-uniform, infinite etc.), since only its intensity participates in the above expression for a potential energy density, the formula should be valid.
In particular, it's valid for an electrical component of an oscillating electromagnetic field.

## Wednesday, October 19, 2022

### Problems 2 on Trigonometric Definitions: UNIZOR.COM - Math4Teens - Trigonometry - Trigonometric Definitions

Notes to a video lecture on http://www.unizor.com

Trigonometry - More Problems on Definitions

1. Prove that for any angle φ the following equality is true
2·(sin6φ + cos6φ) + 1 =
= 3·(sin4φ + cos4φ)

2. Prove that for any angle φ the following equality is true
sin8φ − cos8φ =
= 4·sin6φ−6·sin4φ+4·sin2φ-1

3. Calculate a sum of squares of sines of all angles between 0° and 360° whose measure is a multiple of 30°.

4. Given a point A on a unit circle with positive abscissa xA and positive ordinate yA. Consider a point B with an abscissa equal to an ordinate of A and an ordinate equal to an abscissa of A.
Prove that a sum of angles represented by points A and B is π/2.

5. Solve an equation sin(x)=cos(x).

6. Solve an equation tan(x)=cot(x).