How Does Deriving V = dW/dq Explain Voltage in Circuit Theory?

[PhDeezNutz]

Homework Statement

In my book (Electric Circuits by Nilsson and Riedel) it states

"Whenever a positive and negative charges are separated, energy is expended). Voltage is the energy per unit charge created by the separation. We express this ratio in differential form

$V = \frac{dw}{dq}$"

$v$ = voltage in volts
$w$ = energy in joules
$q$ = charge in coloumbs

It also says a few pages back that one of the assumptions of circuit-theory is

"The net charge on every component in the system is always zero. Thus no component can collect a net excess of charge, although some components, as you will learn later, can hold equal but opposite separated charges"



I'd like to derive the equation for $V$

Relevant Equations

See above

The statement

"The net charge on every component in the system is always zero. Thus no component can collect a net excess of charge, although some components, as you will learn later, can hold equal but opposite separated charges"

leads to believe that we are always dealing with charges of equal magnitude but opposite sign in circuit theory so that is how I will proceed.

Suppose we have two equal but opposite charges

$F = -\frac{q^2}{4 \pi \epsilon_0} \frac{1}{r^2}$

Work done in separating charges (by some external influence)(minus sign because the it opposes the field)(i.e. energy lost)

$W = - \int F \, dr = -\frac{q^2}{4 \pi \epsilon_0} \frac{1}{r}$

Now when it says $V = \frac{dW}{dq}$ I get

$V = - \frac{q}{2 \pi \epsilon_0} \frac{1}{r}$This goes against my intuition (possibly wrong intuition) that the actual expression for $V$ is

$V = \frac{q}{4 \pi \epsilon_0} \frac{1}{r}$

Where did I go wrong?

 

[collinsmark]

Homework Statement:: In my book (Electric Circuits by Nilsson and Riedel) it states

"Whenever a positive and negative charges are separated, energy is expended). Voltage is the energy per unit charge created by the separation. We express this ratio in differential form

$V = \frac{dw}{dq}$"

$v$ = voltage in volts
$w$ = energy in joules
$q$ = charge in coloumbs

It also says a few pages back that one of the assumptions of circuit-theory is

"The net charge on every component in the system is always zero. Thus no component can collect a net excess of charge, although some components, as you will learn later, can hold equal but opposite separated charges"
I'd like to derive the equation for $V$
Relevant Equations:: See above

The statement

"The net charge on every component in the system is always zero. Thus no component can collect a net excess of charge, although some components, as you will learn later, can hold equal but opposite separated charges"

leads to believe that we are always dealing with charges of equal magnitude but opposite sign in circuit theory so that is how I will proceed.

Suppose we have two equal but opposite charges

$F = -\frac{q^2}{4 \pi \epsilon_0} \frac{1}{r^2}$

Work done in separating charges (by some external influence)(minus sign because the it opposes the field)(i.e. energy lost)

$W = - \int F \, dr = -\frac{q^2}{4 \pi \epsilon_0} \frac{1}{r}$

Now when it says $V = \frac{dW}{dq}$ I get

$V = - \frac{q}{2 \pi \epsilon_0} \frac{1}{r}$This goes against my intuition (possibly wrong intuition) that the actual expression for $V$ is

$V = \frac{q}{4 \pi \epsilon_0} \frac{1}{r}$

Where did I go wrong?

Don't overthink that.

For circuit theory (lumped parameter circuits), it's true that everything is assumed to be neutrally charged overall. But lumped parameter systems make a lot of assumptions.

In contrast, it is not necessary to be overall electrically neutral when using electromagnetic theory. (I'm making a distiction between circuity theory and electromagnetic theory here.) Yes, it's true that our universe seems to be electrically neutral on large enough scales, but the theory doesn't require that. If it helps, pretend the system is neutral, but the opposite charge is an infinite distance away, and thus negligible.

When analyzing electromagnetic systems, use a hypothetical "test charge." A test charge has an arbitrary charge magnitude, q, and is always assumed to be positive.

Having negative potentials can arise, and are expected, if takes work to move a positive test charge to zero potential (i.e., "to move the positve test charge out of the potential well.")

 

[scompi]

Generally,

$F = \frac{q_1q_2}{4 \pi \epsilon_0} \frac{1}{r^2}$

Try again using the general form ($q_1q_2$ not $q²$).

 

[PhDeezNutz]

Generally,

$F = \frac{q_1q_2}{4 \pi \epsilon_0} \frac{1}{r^2}$

Try again using the general form ($q_1q_2$ not $q²$).

So we always take the derivative with respect to the test charge (as opposed to the source charge)?

 

[scompi]

So we always take the derivative with respect to the test charge (as opposed to the source charge)?

Yes. If interested, I would suggest going deeper into electrostatics theory. In particular, the definitions of electric potential and electric potential difference (voltage).

 

[vcsharp2003]

Homework Statement:: Work done in separating charges (by some external influence)(minus sign because the it opposes the field)(i.e. energy lost)

Work done by external force would be positive in this case since the displacement and external force vectors are having the same direction. Remember that work done by a force is the dot product of this force vector and resulting displacement vector i.e. $W = \vec F \cdot \vec s=| \vec F | | \vec s | \cos \theta$. If $\vec F$ and $\vec s$ are in same direction then $\theta=0$. This should explain why negative sign in your derivation is not correct. In your derivation $dW= \vec F \cdot \vec {dr}$; this dot product needs to be integrated to find $W$, the total work done.

When unlike charges are being separated, then consider that only one charge is being separated from other charges in system; clearly, the external force must be opposite in direction to attractive force on charge being moved and in same direction as displacement of this charge.

 

[vcsharp2003]

Homework Statement::
Work done in separating charges (by some external influence)(minus sign because the it opposes the field)(i.e. energy lost)

When positive work is done by a force on a system, then the system will gain energy. In this case, since external force vector and displacement vector are in same direction so the work done by external force is positive and therefore, the system gains energy.

 

[Steve4Physics]

Given the thread's title ('Deriving V = dW/dq for a circuit'), it might be worth stepping back for moment.

Your textbook is generalising the formula $V = \frac w q$ to the differential form $V = \frac {dw}{dq}$. Both of these formulae express definitions of voltage (or, more correctly, potential difference). So these formulae can’t be derived.

$V = \frac {dw}{dq}$, essentially means this: when a charge (dq) is moved between 2 points (say A and B) some work (dw) is done. $\frac {dw}{dq}$ is called the voltage (potential difference) between A and B.

If A (say) is taken as the reference point (0V) then V is the potential of B.

It’s worth noting that in the context of circuits, the symbol ‘w’ (or 'W') is often loosely used for any type of energy transfer, e.g. heat dissipated in a resistor.

Here’s a trivial circuit example: when 3C of charge pass (at a constant rate) through a bulb 6J of energy are dissipated as heat and light. The ‘voltage’ across the bulb while the charge flows is $V = \frac w q = \frac {6J}{3C} =2J/C$ (= 2volts because a ‘volt’ means 1 joule/coulomb).

For circuits, we have moving ‘continuous’ streams of charge and internal electric fields. For (simple) electrostatics we have fixed point charges in free space and their radial electric fields. Analysing circuit-voltages in terms of simple electrostatics may not be productive!