When you get shocked from touching a metal doorknob after walking across a rug, what you feel are electrons jumping from your hand to the door metal. The electrons jump because they're attracted to the door by their electric charge. Measures of charge are commonly symbolized as \(q\), and charge is a characteristic property of all matter.

The standard unit of charge in the SI system is the Coulomb. Some types of matter, like neutrons, have no charge and are electrically neutral (hence their name). Other particles like protons and electrons, have positive or negative electric charge, respectively. Despite the unit of charge being the Coulomb, the "unit" particles of charge, the proton and the electron, each have a charge of \(\pm 1.6\times10^{-19}\text{ C}.\)

The SI unit of charge is the Coulomb, abbreviated as \(\text{C}:\)

\[1\textrm{ C}\approx 6.25\times 10^{18} q_e.\]

Particle colliders have revealed particles called quarks, which come together in particular combinations to form other particles such as the neutron and proton, and they exhibit charges of \(\{+\frac23 q_p, +\frac13 q_p, -\frac13q_p, -\frac23 q_p\}\).

They are more fundamental than protons and neutrons, but because of their strong tendency to join up into more stable aggregate particles, they are rarely seen outside the special situation of the particle collider.

The proton consists of three quarks (

up,up, anddown) of charge \(\{+\frac23 q_p, +\frac23 q_p, -\frac13 q_p\}\) so that\[q_p = \left(\frac23+\frac23-\frac13\right)q_p=q_p\] as expected.

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Charges wouldn't be all that interesting if they didn't enable some kind of motion. In fact, charged matter is seen to undergo peculiar phenomena that are not experienced by uncharged matter. Some examples of these are the well-known **aurora borealis**, electrons flowing *en masse* from cloud to cloud during storms (a.k.a. lightning), and when your hair stands on end after riding down a slide.

One of the simplest interactions that a charged particle can have is with an electric field. The electric field is essentially a 3D grid that fills all of space, and records a value and direction at every point corresponding to the force that a charged particle would experience if it were placed at that point. Hence, if a positively charged particle is in an electric field, it experiences a push along the local direction of the field while a negatively charged particle will experience a push along the direction opposite the local direction of the field. This is an important definition which should be noted in problem solving.

An electric field \(\vec{E}(\vec{r})\) is defined at every point in space \(\vec{r}\), and acts on positively charged particles in the direction \(\frac{\vec{E}}{\left|E\right|}\). Similarly, it acts upon negatively charged particles in the direction \(-\frac{\vec{E}}{\left|E\right|}\).

In two dimensions, we can visualize the electric field as a lattice of arrows:

The length and width of each arrow corresponds to the field strength at the underlying point. While this representation is discrete, the electric field is continuous, and therefore can be interpolated between the arrows shown in the grid.

Mathematically, we have

\[\begin{align}\vec{F} &= q\vec{E} \\&= m\vec{a}.\end{align}\]

See Also19.1 Electric Potential Energy: Potential Difference - College Physics 2e | OpenStax19.2 Electric Potential in a Uniform Electric Field - College Physics 2e | OpenStaxDifference between F=qE and F=q(V*B)18.4: Electric Field- Concept of a Field RevisitedThis relation is a special case of the

Lorentz force lawwith \(\vec{B}=0\) (when magnetic fields are present, there is an extra term that is beyond the focus of this article).

For the purposes of classical electrodynamics, an electric field is more or less defined by this relationship in that we can put a charge at various locations, measure the force it feels, and use the Lorentz force law to calculate the electric field at each point.

Stop that muonA high energy muon (of charge \(-q_p\)) enters the upper atmosphere along a trajectory straight through the center of the Earth. There are two very large clouds (one directly above the other) separated by a distance of \(l=100\text{ km}\), with an electric field of constant strength between them, that is aligned vertically.

If the muon has an incoming kinetic energy of \(\text{KE}_i = 6.4\times10^{-16}\text{ kJ}\), how strong would the electric field between the two clouds have to be in order to bring the muon to rest before it passes the bottom cloud? (Ignore breakdown reactions and assume the muon to be a stable particle.)

In the electric field, the muon will experience a force of \(F=q_pE\). At most, the muon can travel \(100\text{ km}\) before coming to rest, before which the field will have performed work in the amount \(W=F\cdot d = q_pEl\) on the muon.

Therefore, we can say that \(W = \text{KE}_i\), and the minimal electric field strength is given by \[E = \frac{\text{KE}_i}{q_pl} \approx 40\textrm{ N/C}. \]

A circular ring of radius \(a\) is uniformly charged with charge \(+Q\text{ C}\) at its circumference. Find the electric field acting at a point distance \(x\) from the center of the ring.

The figure clearly illustrates the idea. Let the point be \(P\) at a distance of \(x\) units from the center of the circular ring \(A\) with radius \(a\) units.

The electric field acting \(\vec{E}\) on \(P\) along \(\vec{DF}\) making angle \(\theta\) with the horizontal can be divided into two components, namely \(E\cos\theta\) and \(E\sin\theta\).

The vertical component however doesn't contribute to the electric field of the point, because for any two elementary portions of the circular loop that are opposite to each other and equal will cancel. So the net electric field will be towards the \(x\)-axis.

The distance of the point from the ring is \(\sqrt{x^2+a^2}.\) Now consider an elementary length \(dl\) of the loop, the charge on which is given by \(dq=\frac{q}{2\pi a}dl.\) Thus,

\[\begin{align}dE|\vec{dE}|&= \frac{dq}{4\pi{\varepsilon}_{0}}\frac{1}{\left(\sqrt{x^2+a^2}\right)^2}\\dE_x&=dE\cos\theta \\&= dE \cdot \frac{x}{(x^2+a^2)^{1/2}} \\&= \frac{1}{4\pi{\varepsilon}_{0}}\frac{q\cdot x\cdot dl}{2\pi a(x^2+a^2)^{3/2}},\end{align}\]

as \(\cos\theta\) is derived from the right triangle \(ADP.\)

Now, the electric field due to the entire circular loop is given by

\[\begin{align}E&=\int_{whole loop}^{} dE_x \\&= \int_{whole loop}^{} \frac{1}{4\pi{\varepsilon}_{0}}\frac{q\cdot x\cdot dl}{2\pi a(x^2+a^2)^{3/2}}\\&= \frac{1}{4\pi{\varepsilon}_{0}}\frac{q\cdot x}{2\pi a(x^2+a^2)^{3/2}} \int_{whole loop}^{}dl\\&=\frac{1}{4\pi{\varepsilon}_{0}}\frac{q\cdot x\cdot (2\pi a)}{2\pi a(x^2+a^2)^{3/2}} \qquad \text{(the entire length of the loop is } 2\pi a) \\&=\frac{1}{4\pi{\varepsilon}_{0}}\frac{q\cdot x}{(x^2+a^2)^{3/2}}.\end{align}\]

Some special cases:

\(P\) lies at the center of the loop: in this case \(x=0\) and thus \(E=0.\)

\(P\) is such that \(x>>a,\) so \(a^2\) can be neglected in comparison to \(x^2\): in this case \(E=\frac{1}{4\pi{\varepsilon}_{0}}\frac{q}{x^2}.\)

In addition to responding to external electric fields, charged particles give rise to electric fields of their own. Knowing the behavior of these particle fields allows us to build up our understanding to more complicated arrangements of charge, which are crucial in applying our knowledge to engineering problems.

As such, experimental physicists of the \(18^\text{th}\) century like Priestley and Coulomb carefully measured the electric field of charges in the lab. They used clever arrangements to measure the force felt by an object of charge \(q\) (**test charge**) when placed in the vicinity of another object of charge \(Q\) (**source charge**). One way to do this is to hold \(Q\) fixed (e.g. mount it on an electrically neutral stick) and measure the force felt by \(q\) as it is placed at various locations within the field of \(Q\). After measuring enough points, an approximate picture emerged like the arrow diagram below.

One thing they found is that the force felt by \(q\) was independent of direction, i.e. the force is spherically symmetric and points directly along the vector connecting \(q\) with \(Q\). In other words, \(\vec{F}(\vec{r}) = \vec{F}(r)\). Another was a strong dependence on the distance away from \(Q\). The figure below shows data for the relative strength of the force felt by \(q\) as a function of the distance from \(Q\), as well as the curves \(r^\alpha\) for \(\alpha = 1,-1,-2,-3\).

It is clear that the force decreases with the inverse square of the distance, and therefore \(F \propto \frac{1}{r^2}\). The force was also found to increase linearly as \(q\) or \(Q\) increased, and therefore \(F\propto qQ\). Finally, the force was attractive if \(q\) and \(Q\) had opposite signs, and it was repulsive for like sign.

Putting all these observations together, we can say \(F \sim \frac{qQ}{r^2}\). According to the Lorentz force law, the force felt by the test charge is given by \(F=qE_Q\), hence \(F=qE_Q\sim\frac{qQ}{r^2}\) and \(E_Q\sim \frac{Q}{r^2}\), and thus we have

\[E_Q=k\frac{Q}{r^2},\]

where \(k\) is a constant of proportionality that has been determined experimentally as \(9\times 10^9\text{ Nm}^2\text{/C}^2\). This is what's known as Coulomb's law.

Coulomb's lawThe strength of the electric field at position \(r\) due to a point charge \(Q\), is given by \[E = k\frac{Q}{r^2}.\]The field points along the vector from position \(r\) to the charge. By convention, field lines point inward toward negatively charged particles like electrons and outward for positively charged particles like protons.

While Coulomb's law is strictly true only for point charges, it is still an excellent approximation for the electric field far away from more complicated arrangements of particles. Close-up, an arbitrary arrangement of charges can have an electric field of great detail that isn't easily visualized without a computer.

However, as the scale increases, these local variations will drop off quickly, and at great distances the field will tend to look like \[\displaystyle E(x) = k \sum_i\frac{q_i}{(r_i-x)^2} \approx k\frac{Q_\text{net}}{(\bar{r}-x)^2}.\]

If \(Q_\text{net}>0\), the field lines will point radially outward, and if \(Q_\text{net}<0\), the field lines will point radially inward toward the center of the distribution.

Twin peaks

Two charges of strength \(q\) are placed at \(r_-=-\epsilon\) and \(r_+=+\epsilon\). The field close to either charge is given by \[E(r) = k\frac{q}{\left(r+\epsilon\right)^2}+k\frac{q}{\left(r-\epsilon\right)^2}.\] What does the field look like at very far distances?

When \(r \gg \epsilon\), we can approximate \(E(r)\) as \(2k\frac{q}{r^2} + 6k\frac{q\epsilon^2}{r^4}\). Because \(\epsilon^2/r^2\approx 0\), we can ignore the second term.

Thus, the field looks like \(\displaystyle 2k\frac{q}{r^2}\) far away from the charges, which is equal to \(\displaystyle k\frac{Q_\text{net}}{r^2}\).

Below, we show the same dipole field as seen zoomed out by a factor of ten. The local structure of the field around the dipoles is no longer visible, and the arrangement looks roughly the same as the field for one charge of strength \(2q\).

\[\frac{3}{2}d\] \[2d\] \[\frac{5}{2}d\] \[3d\]

A particle with electric charge \(-q\) enters a uniform electric field at the point \(P=(0, 3d).\) The direction of the electric field is the \(+y\) direction. The charged particle moves along a projectile path inside the electric field. After exiting the electric field, it shows a uniform motion, arriving at \(Q=(4d, 0).\) If another charged particle with the same mass but a different electric charge of \(-2q\) enters the electric field in the same way as above, what will be the destination point on the \(x\)-axis?

Ignore the gravitational force and the sizes of the charged particles.

In the last example, we exploited a property of electric fields called superposition. The principle of superposition states that in the presence of multiple sources of electric field, the resultant field is simply the sum of the individual fields at each point.

Principle of superpositionIn the presence of multiple fields \(E_1(r),E_2(r),\ldots, E_n(r)\), the field strength at \(r\) is given by

\[E_\text{tot}(r) = \sum_i E_i(r).\]

Near or far, flux remains.Coulomb's law suggests a curious quantity that ought to be the same for any spherical surface centered on a particle. If we multiply the strength of the electric field everywhere on the surface, by the surface area of the small patch of surface that the field penetrates, we get

\[\begin{align} \Phi_E &= \sum E(r_i) \times \Delta A(r_i) \\ &= k\frac{q}{r^2}\times 4\pi r^2 \\ &= 4\pi k q. \end{align}\]

Regardless of how big or how little the encapsulating sphere is made to be, this quantity, the field everywhere on the surface, times the surface area, will always be equal to \(4\pi k\) times the total charge enclosed by the surface.

In fact, the surface doesn't need to be centered on the charge, and the surface need not be spherical. The relation \(\Phi_E = 4\pi k Q_\text{enc}\) holds for any closed surface whatsoever that encloses the charge \(Q_\text{enc}\).

This is a rather curious observation. Could this flux coincidence have fundamental importance?

To close, we'll compare the strength of gravity with the strength of the Coulomb force.

## Comparison with gravity

\[\] To make the comparison fair, we'll compare the gravitational attraction of two protons with their electric repulsion. The gravitational attraction between them is given by \(G\frac{m_p^2}{r^2}\), so

\[\begin{align} F_E/F_G &= k\frac{q_p^2}{r^2} \frac{1}{G}\frac{r^2}{m_p^2} \\ &= \frac{kq_p^2}{Gm_p^2} \\ &\approx \frac{9\times 10^9}{6.6\times 10^{-11}}\frac{\left(1.6\times^{-19}\right)^2}{\left(1.7\times10^{-27}\right)^2} \\ &\approx 10^{36}. \end{align}\]

It is safe to say that the gravitational attraction of two protons is utterly insignificant next to their electric repulsion.

Given this enormous discrepancy, we might wonder when, if ever, gravitational interactions are worth considering. The saving grace for gravity is that because Coulomb forces are so strong that charged objects tend to pair up in such a way that most macroscopic objects are electrically neutral. As charge-neutral objects don't participate in Coulomb interactions, there is little Coulomb force to speak of.

On the other hand, gravity has no "negative particle," so all masses contribute to the attractive force. For very massive objects gravitational interactions can be quite significant.