Consider two fixed point charges, $q_{1}$ and $q_{2}$, separated by a distance $r$. Coulomb discovered experimentally that the electrostatic force $F$ on charge $q_{2}$ due to its interaction with $q_{1}$ is given by the equation

(1) $F = k \frac{q_{1} q_{2}}{r^{2}}$ Since $F$, $q$, and $r$ are all defined, $k$ becomes a constant of proportionality which must be determined experimentally. Equation (1) is now known as Coulomb's law.

In the SI system, the unit of charge $q$ is the coulomb. It is defined as the amount of charge which passes a point in a circuit when one ampere flows for one second. Stated mathematically,

(2) $q = \int Idt$ or $I = \frac{dq}{dt}$

Coulomb's law is the fundamental law of electrostatics. However, when one considers the forces acting between moving charges, one finds that Coulomb's law is not the most convenient form for making calculations. The equation is, therefore, reformulated into the much more general set of equations known as Maxwell's equations. In order to simplify Maxwell's equations, it is convenient to write the constant 𝑘 in another form, i.e., $k = \frac{1}{4 \pi \epsilon_{o}}$. Thus Coulomb's law can also be written as

(3) $F = \frac{1}{4 \pi \epsilon_{o}} \frac{q_{1} q_{2}}{r^{2}}$

The constant $\epsilon_{o}$ is called the permittivity of free space.

Equation (3) applies to point charges. Unfortunately, in the laboratory it is not possible to work with true point charges. Therefore, we must reformulate Coulomb's law so that it applies to a practical experiment. One such practical experiment is measurement of the force between two parallel plates maintained at a known potential difference. We shall use this approach in this experiment. We must, therefore, reformulate Coulomb's law so that it applies to this arrangement.

You will recall that in mechanics we found that the gravitational force $F$ acting on a mass $m$ was given by Newton’s second law,

(4) $\vec{F} = m \vec{g}$

where $\vec{g}$ is the acceleration due to gravity. The concept of a gravitational field was very useful in calculations even though it obscured the fundamental fact that a gravitational force results from the interaction of two or more masses.

By analogy to the gravitational case, Eq.(4), it should be clear that it is useful to define an electric field by the equation

(5) $\vec{F} = q \vec{E}$

Comparison with Coulomb's law, Eq.(3), shows that for a point charge $q_{1}$,

(6) $\vec{E} = \frac{1}{4 \pi \epsilon_{o}} \frac{q_{1}}{r^{2}_{1}}bf \vec{r_{1}}$

One could calculate $\vec{E}$ for a large flat plate by setting up Eq.(6) in differential form for an element of the total charge on the plate, $dq$, and then integrating over the whole plate. It is, however, more convenient to recast Coulomb's law in still another form. This relation is known as Gauss' law. Gauss' law uses the concept of electric flux.

The electric flux $d\phi$ through an element of surface of area $ds$ is defined as the dot product of the electric field strength $vec{E}$ and the vector pointing normal to the infinitesimal area $ds$. (See Fig. 1.)

Figure 1: Definition of electric flux

Mathematically, this statement is

(7) $d\phi = \vec{E} \cdot d \vec{s}$

or, in simplified form,

(8) $d\phi = E cos(\theta) ds$

Thus $d\phi$ is the amount of flux of electric field passing through the area $ds$.

(The concept of flux can perhaps be more easily understood if one considers another vector field, e.g. the velocity of a fluid. In this case $\vec{v} \cdot ds$ is the amount of fluid flowing through the area $ds$.)

Consider now a point charge $q$ and let us calculate the total flux through a closed surface surrounding that charge. For convenience we take the closed surface as a sphere centered on the charge as shown in Fig. 2.

In this particular case, $\vec{𝐸}$ and $\vec{𝑑𝑠}$ are parallel. Thus $\theta = 0^{\circ}$ and eq.(7) can be rewritten

(9) $d \phi = E ds$.

Noting that $E$ is a function of $r$ and integrating over the complete sphere, we have

(10) $ \phi = \int_{S} 𝐸(𝑟)𝑑𝑠 $.

By symmetry, we can be sure that $E$ is constant at any particular radius. Therefore, it can be brought out in front of the integral. The integral is now simply the total surface area of the sphere. Therefore, it can be brought out in front of the integral. The integral is now simply the total surface area of the sphere. Therefore,

(11) $\phi = E(r) 4 \pi \epsilon_{\circ}$.

Combining this with the definition of $E$ at radius $r$, i.e., the scalar form of eq.(6), we have

(12) $ \phi = \frac{1}{4 \pi \epsilon_{\circ}} \frac{q}{r^{2}} 4 \pi r^{2} = \frac{q}{\epsilon_{\circ}}$.

We have proved eq.(12) only for a spherical surface surrounding a point charge. Application of the superposition theorem and the concept of solid angle shows that the theorem applies to the total charge inside that closed surface and that the integral is zero for all charges outside that surface. Stated mathematically, Gauss' law is

(13) $ \int _{S} \vec{𝐸}(𝑟) \cdot 𝑑\vec{𝑠} = \frac{q_{i}}{\epsilon_{\circ}}$.

where $𝑞_{𝑖}$ includes only those charges inside the gaussian surface.

Consider a parallel plate capacitor consisting of two large conducting plates of area $A$ separated by a small distance $d$. The plates carry equal and opposite charge $q$ and we ground the negatively charged plate. Let us first calculate the electric field between the plates using Gauss's law. As shown in Fig. 3, one end of the Gaussian “pillbox” lies inside the conductor where there is no field, while the other end lies between the plates in the field which we wish to determine.

By eq.(13),

(14) $\int E ds = ES = \frac{\sigma S}{\epsilon_{\circ}}$.

Solving for $E$, we have

(15) $E = \frac{\sigma}{\epsilon_{\circ}}$.

Note that the magnitude of the charge, $∣𝑞∣$, on either plate is $\sigma A$. Since it is easier to measure the voltage $V$ between the plates than to measure $q$, we need a relationship between $V$ and $q$. The work done in transporting a charge $\delta q$ from the grounded plate to the positively charged plate which at potential $V$ is $\delta W$, i.e.

(16) $ \delta W = \delta q V$.

Work is defined mechanically as force times distance,

(17) $ \delta W = \delta q E d$.

Therefore, the relation between voltage and field is

(18) $E = \frac{V}{d}$,

and the relation between voltage and charge is

(19) $ V = \frac{\sigma d}{\epsilon_{\circ}} = \frac{q d}{A \epsilon_{\circ}}$.

It is useful to rewrite eq.(19) as

(20) $V = \frac{q}{C}$,

where $C$ is called the capacitance and depends only on the geometry of the plates (i.e., their area and separation). For a parallel plate capacitor, the capacitance is given by

(21) $ C = \frac{A \epsilon_{\circ}}{d}$.

The unit of capacitance is the farad.

Finally, we will need to know the energy $U$ stored in a capacitor. The energy stored is the work required to place a charge $q$ onto the capacitor plates. From eq.(21),

(22) $dU = V dq = \frac{q dq}{C}$.

Therefore,

(23) $ U = \int \frac{q}{C} dq$.

Combining eqs.(21) and (23) we can also write

(24) $U = \frac{1}{2} q V$.

We will find the force $F$ between the plates by conducting a “thought experiment” in which the plates’ separation is decreased by an amount $\delta d$ while a battery maintains a constant voltage $V$ between the plates. Let us see how energy is conserved in this process.

The mechanical work performed by the plates is

(25) $ \delta W_{plates} = -F \delta d$.

Notice that $\delta W_{plates}$ is positive since $\delta d$ is negative and $F$ is positive. The work done by the battery as the charge on the plates is increased by an amount $\delta q$ to maintain a constant voltage is

(26) $\delta W_{batt} = V \delta q$.

However, we cannot simply equate the two terms because the energy $U$ of the capacitor also changes. According to eq.(24),

(27) $\delta U = \frac{1}{2} q \delta V$.

Therefore, by conservation of energy,

(28) $V \delta q = -F \delta d + \frac{1}{2} V \delta V$,

so that

(29) $F \delta = -\frac{1}{2} V \delta q$

or

(30) $F = -\frac{1}{2} V \frac{\delta d}{\delta q}$.

From eq.(19), the product $qd$ is constant at constant voltage, so that

(31) $\frac{\delta q}{\delta d} = -\frac{q}{d}$,

and we obtain our formula for the force between the plates:

(32) $F = \frac{1}{2} V \frac{q}{d}$.

From eq.(19) this can also be expressed as

(33) $F = \frac{\epsilon_{\circ} A V^{2}}{2 d^{2}}$.