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Electric potential as work per charge, $V = kq/r$ added as scalars, the field as its negative gradient, equipotentials, and energy in electron volts.
Paper packet. Every task here also exists on screen, where it is checked automatically; answers written on paper are not assessed by Nydus. When you are back at a device, enter your answers there.
By the end of this lesson you will be able to find potentials from charges and fields, fields from potentials, and energies of charges moving through potential differences.
From the mechanics lessons you know that a conservative force has a potential energy with $F = -dU/dx$. From the last two lessons you can find electric fields. Here the two ideas combine: the electric force is conservative, so it has a potential energy, and dividing by charge gives a potential that belongs to the field alone.
| Term | What it means |
|---|---|
| Electric potential | $V = U/q$, potential energy per unit charge, in volts (J/C). |
| Potential difference | $V_B - V_A = -\int_A^B \vec{E}\cdot d\vec{l}$; the voltage between two points. |
| Potential gradient | The rate of change of $V$ with position; $\vec{E} = -\nabla V$. |
| Equipotential | A surface on which $V$ is constant; field lines cross it at right angles. |
| Electron volt | $1$ eV $= 1.60 \times 10^{-19}$ J, the energy an electron gains across $1$ V. |
| Reference point | Where $V = 0$ is chosen, usually infinitely far away. |
| Volt per meter | The unit V/m, identical to N/C. |
The electric force is conservative, so moving a charge from $A$ to $B$ takes the same work on every path. The potential is that work per unit charge:
$$V_B - V_A = -\int_A^B \vec{E}\cdot d\vec{l}.$$
Going the other way, the field is the negative slope of the potential:
$$E_x = -\frac{\partial V}{\partial x}, \qquad \vec{E} = -\nabla V.$$
For a point charge, with $V = 0$ at infinity, $V = kq/r$. Potential is a scalar, so the potential of several charges is the plain sum of theirs, with signs but no components. That makes it the easy route to fields: find $V$ by adding, then differentiate. A charge $q$ moving through a potential difference $\Delta V$ changes its potential energy by $q\Delta V$.
Another way: picture
Picture a contour map of hills. Height is the potential; the contour lines are equipotentials. The steepest way downhill, straight across the contours, is the field's direction, and the steepness is its strength. Where contours crowd together the field is strong. A positive charge rolls downhill, a negative charge up.
Another way: steps
The potential difference between two points is minus the line integral of the field. In a uniform field $E$ along $+x$, $V(x) = V_0 - Ex$: the potential falls steadily in the field's direction, and a potential difference $\Delta V$ over a distance $d$ means a field $\Delta V/d$. This is why the units V/m and N/C are the same.
For a point charge, integrating $E = kq/r^2$ from infinity inward gives $V = kq/r$. Outside a charged sphere the same result holds; inside a conducting sphere, where $E = 0$, the potential is constant at its surface value $kQ/R$. Potential is continuous even where the field jumps.
If you know $V$ everywhere, the field is its negative gradient: $E_x = -\partial V/\partial x$, and likewise for $y$ and $z$. For $V = kq/r$, $E_r = -dV/dr = kq/r^2$, recovering Coulomb's law. For $V = 3x^3 - 5x$, $E_x = 5 - 9x^2$.
The gradient points uphill, so the minus sign makes the field point downhill. Where $V$ has a maximum or minimum, its slope is zero, so the field is zero, even if $V$ itself is large. At the center of a charged ring, $V = kQ/R$ is at its highest on the axis, and $E = 0$.
Because potential is a scalar, adding potentials needs no geometry beyond distances. Two charges $+q$ and $-q$ at equal distances from a point give $V = 0$ there, though the field there is not zero. A ring gives $V = kQ/\sqrt{z^2 + R^2}$ on its axis, with no need to cancel components, since every element is at the same distance.
Differentiating the ring's potential with respect to $z$ gives $E_z = kQz/(z^2 + R^2)^{3/2}$, the result found by vector integration two lessons ago. The route through potential is usually shorter: one scalar integral, then a derivative.
Surfaces of constant $V$ are equipotentials. Moving a charge along one takes no work, so the field has no component along it: field lines always cross equipotentials at right angles. Around a point charge they are spheres; in a uniform field, parallel planes.
A conductor in equilibrium has no field inside, so its entire volume, surface included, is one equipotential. That is why a wire connecting two conductors brings them to the same potential, and why the field just outside a conductor meets it perpendicularly. Near sharp points, equipotentials crowd together, making the field strong.
A charge $q$ moving through $\Delta V$ gains kinetic energy $-q\Delta V$ if only the electric force acts. A proton falling through $1000$ V gains $1.6 \times 10^{-16}$ J; an electron moving from low to high potential gains the same. For particles, the natural unit is the electron volt, the energy an electron gains across $1$ V: $1.60 \times 10^{-19}$ J.
Energies of chemical bonds are a few eV, visible light photons about $2$ eV, X-rays thousands of eV. Particle accelerators quote energies in MeV, GeV and TeV; the Large Hadron Collider gives protons several TeV each.
Checking an answer. $V$ from positive charge must be positive and fall with distance. $\vec{E}$ must point from high to low $V$. The derivative of your $V$ must reproduce any field you know independently. And energy conservation must hold: a charge released from rest moves so as to lower its potential energy.
Defining a potential is possible only because the work done by the electric force does not depend on the path. That is true for static charges, because the inverse-square force is radial; around each point charge, any path can be broken into radial pieces, which contribute, and pieces along spheres, which do not.
Adding potentials rests on superposition of fields: if fields add, their line integrals add. Choosing where $V = 0$ is free, since only differences are physical, but the choice must be the same for every charge in a sum. For an infinite line or sheet the potential at infinity is not finite, so a nearby reference is chosen instead.
A battery maintains a fixed potential difference between its terminals: $1.5$ V for an AA cell, $12$ V for a car battery, $120$ V for a household outlet in the United States. Charges moving through a circuit lose that potential energy in the components, as heat in a resistor or as light in a bulb.
Every wire in a circuit diagram is ideally an equipotential, since good conductors need almost no field to carry current. Voltage in circuits is exactly the potential difference of this lesson, and the next lessons on capacitors and circuits use it throughout. A voltmeter, whatever its design, measures exactly this: the work per unit charge between the two points its leads touch.
The potential energy of a pair of point charges is $U = kq_1q_2/r$: the work needed to bring them together from far apart. It is positive for like charges, which must be pushed together, and negative for unlike charges, which pull together on their own. For three or more charges, add the energy of every pair once: three charges have three pair terms, four have six.
This gives the energy stored in an arrangement, and so the work to assemble or dismantle it. The energy of a hydrogen atom's electron and proton at $0.053$ nm apart is $-27$ eV; half of that is taken back as the electron's kinetic energy, leaving the familiar binding energy of $13.6$ eV. The same pair sum, over the billions of charges in a molecule, is what chemists compute to predict how molecules hold together and react. A negative total means the group is bound: energy must be supplied to pull it apart, and it is released when the group forms.
A useful check on any such sum is to rebuild the arrangement one charge at a time. The first charge costs nothing; each later one costs its charge times the potential already made by those before it. The total is the same whichever order is chosen, a direct consequence of the force being conservative.
Every nerve cell keeps its inside about $70$ mV negative relative to the outside, pumping ions across a membrane only about $7$ nm thick. The field across the membrane is therefore about $10^7$ V/m, stronger than the field that makes air spark. The membrane survives because it is a thin insulating film of fat.
A nerve signal is a wave of potential that travels along the cell: channels open, ions rush through, and the inside briefly swings to $+30$ mV before recovering. Researchers at the National Institutes of Health and elsewhere measure these potentials with glass electrodes a fraction of a micrometer across, and the potential difference is what drugs for epilepsy, pain and heart rhythm act on.
Old television and oscilloscope screens made pictures by accelerating electrons through tens of thousands of volts and steering them onto phosphor. The same electron gun, boiling electrons off a hot filament and accelerating them through a potential difference, is at the heart of electron microscopes and X-ray tubes today.
In a hospital X-ray tube, electrons fall through about $100{,}000$ V and slam into a metal target, where each can produce an X-ray photon of up to $100$ keV. The potential difference sets the maximum photon energy, and so how penetrating the X-rays are: radiologists choose lower voltages for soft tissue and higher ones for bone.
The field is the slope of the potential, not its value. Midway between equal and opposite charges the potential is zero but the field is strong; at the center of a charged ring the potential is at its highest on the axis but the field is zero.
A related error is to add potentials as vectors, taking components. Potential has no direction; only its sign matters. Adding components of $kq/r$ gives nonsense, while adding the values themselves gives the right answer at once.
Charges $+6.0$ nC at $x = 0$ and $-2.0$ nC at $x = 0.40$ m. Find the potential at $x = 0.20$ m.
$V = \dfrac{9 \times 6.0}{0.20} + \dfrac{9 \times (-2.0)}{0.20} = 270 - 90 = 180\ \text{V}$
Scalars with signs.
Find the potential at $x = 0.60$ m.
$V = \dfrac{54}{0.60} - \dfrac{18}{0.20} = 90 - 90 = 0$
Zero potential, but not zero field.
Find the field there.
$E = \dfrac{54}{0.36} - \dfrac{18}{0.040} = 150 - 450 = -300\ \text{N/C}$
Pointing toward $-x$, toward the negative charge.
Find the work to move $1.0$ nC from $x = 0.60$ to $0.20$ m.
$W = 1.0 \times (180 - 0) = 180\ \text{nJ}$
$q\Delta V$.
Check that the path does not matter.
$\text{only the endpoint potentials entered}$
A conservative force.
In a region $V = 4.0x^2 - 2.0y$ volts. Find the component $E_x$.
$E_x = -\dfrac{\partial V}{\partial x} = -8.0x$
Treat $y$ as constant.
Find the component $E_y$.
$E_y = -\dfrac{\partial V}{\partial y} = 2.0\ \text{V/m}$
Treat $x$ as constant.
Evaluate the field at $(1.0, 3.0)$ m.
$\vec{E} = (-8.0, 2.0)\ \text{V/m}$
Components.
Find its magnitude.
$E = \sqrt{64 + 4} = 8.2\ \text{V/m}$
Pythagoras.
Find where $E_x$ vanishes.
$x = 0$
Where $V$ is flat in $x$.
Describe the equipotentials.
$4.0x^2 - 2.0y = \text{const}: \ y = 2.0x^2 + c$
Parabolas, crossed at right angles by field lines.
Two plates $2.0$ cm apart are held at $0$ and $500$ V. Find the field.
$E = \dfrac{500}{0.020} = 2.5 \times 10^4\ \text{V/m}$
From the $500$ V plate toward the $0$ V plate.
An electron leaves the $0$ V plate from rest. Find its energy at the other.
$\Delta K = e\Delta V = 500\ \text{eV} = 8.0 \times 10^{-17}\ \text{J}$
It moves toward higher potential.
Find its speed.
$v = \sqrt{\dfrac{2 \times 8.0 \times 10^{-17}}{9.11 \times 10^{-31}}} = 1.33 \times 10^7\ \text{m/s}$
About 4 percent of light speed.
Find its acceleration.
$a = \dfrac{eE}{m} = \dfrac{1.6 \times 10^{-19} \times 2.5 \times 10^4}{9.11 \times 10^{-31}} = 4.4 \times 10^{15}\ \text{m/s}^2$
Enormous, for a tiny mass.
Find the travel time.
$t = \dfrac{v}{a} = \dfrac{1.33 \times 10^7}{4.4 \times 10^{15}} = 3.0\ \text{ns}$
Constant acceleration.
Check the distance.
$\tfrac{1}{2}at^2 = \tfrac{1}{2}(4.4 \times 10^{15})(3.0 \times 10^{-9})^2 = 0.020\ \text{m}$
The plate spacing.
Repeat for a proton from the $500$ V plate.
$v = 1.33 \times 10^7 \times \sqrt{\dfrac{9.11 \times 10^{-31}}{1.67 \times 10^{-27}}} = 3.1 \times 10^5\ \text{m/s}$
Same energy, far more mass.
Add the potentials.
$V = \dfrac{9 \times 4.0}{0.30} - \dfrac{9 \times 4.0}{0.30} = 0$
Equal and opposite.
Find each charge's field there.
$E_1 = E_2 = \dfrac{36}{0.090} = 400\ \text{N/C}$
Both point toward the negative charge.
Add the fields.
In a region of uniform field, the potential rises steadily by $2$ V over $4$ cm in the $+x$ direction. What is the electric field?
Complete the worked solution: a ring of radius $1.2$ m carries $3$ nC. With $k = 9.0 \times 10^9$ N·m²/C², find the potential at its center and at a point $0.9$ m along its axis, in V, and the work needed to push a $+1.0$ nC charge from that point to the center, in nJ.
Evaluate the potential at the center.
$V_0 = \dfrac{kQ}{R} =$ c
Every element is $R$ away.
Evaluate the potential on the axis.
$V_z = \dfrac{kQ}{\sqrt{z^2 + R^2}} =$ a
No components needed: potential is a scalar.
Multiply the difference by the charge moved.
$W = q'(V_0 - V_z) =$ w
In nC times V, which is nJ.
Compare with the field at the center.
$E_0 = 0 \text{ although } V_0 \neq 0$
The field is the slope; the potential is at a peak there.
Match each quantity to its expression.
| $kq/r$ | $-dV/dx$ | $-\int E_x\,dx$ | $q\Delta V$ | |
|---|---|---|---|---|
| potential of a point charge | ||||
| field from potential | ||||
| potential difference from field | ||||
| potential energy change |
A $+2$ nC point charge is fixed in place. With $k = 9.0 \times 10^9$ N·m²/C², fill in the potential $0.45$ m away in V, the potential $0.9$ m away in V, and the work needed to push a $+2.0$ nC charge from the farther point to the nearer, in nJ.
| value | |
|---|---|
| potential, nearer (V) | |
| potential, farther (V) | |
| work to move inward (nJ) |
Along the $x$-axis the potential in a region is $V = 2x^3 - 3x$ volts, with $x$ in meters. Write the field component $E_x$, in V/m, as a formula in $x$.
Answer:
An electron starts from rest and is accelerated through a potential difference of $400$ V. With $e = 1.60 \times 10^{-19}$ C and $m = 9.11 \times 10^{-31}$ kg, how fast is it moving at the end, in units of $10^6$ m/s?
Answer: × 10⁶ m/s
A neuroscientist at the National Institutes of Health measures a resting potential of $80$ mV across a nerve cell's membrane, which is $5$ nm thick. Treating the field in the membrane as uniform, how strong is it, in MV/m?
Answer: MV/m across the membrane
Lesson test: one question per skill, one attempt each, no hints. Your answers are checked when you submit.
Along the $x$-axis the potential in a region is $V = 2x^3 - 8x$ volts, with $x$ in meters. Write the field component $E_x$, in V/m, as a formula in $x$.
Answer:
You can move between field and potential. Explain to someone how the potential can be zero at a point where the field is strong.
20. Your turn: find the potential midway between $+4.0$ nC and $-4.0$ nC charges $0.60$ m apart, and the field there., step 3
$E = 800\ \text{N/C}$
Zero potential, strong field.