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 P_E 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 Q_max with corresponding difference of potential between the plates (voltage) being equal V_max. 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 V_max can be considered a function of accumulated charge W=W(Q) and it grows from zero to some maximum W_max - the energy accumulated by a capacitor as it accumulated a charge of Q_max.
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 V_max.

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 W_max needed to charge a capacitor from zero to Q_max, we have to integrate dW from 0 to Q_max:
W_max = ∫_[0,Qmax][Q/C]·dQ =
= ½·Q²_max/C

Of course, the amount of charge Q_max 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 = ε_r·ε₀
where ε_r 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 P_E 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
P_E = 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.

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 x_A and positive ordinate y_A. 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).