Unit 3

Electric Potential

Units 1 and 2 described electricity with a vector field: an arrow at every point. This unit replaces the arrows with a single number at every point, the electric potential \(V\), measured in volts. Think of it as a height map: charges roll “downhill” in potential the way a ball rolls downhill in a gravitational landscape. Every voltage you will ever measure with a multimeter, every node voltage in a circuit simulation, every “3.3 V logic” rail is a statement about this one quantity. It is also easier to compute than the field (numbers add, arrows do not), and once you have \(V\) you get \(\mathbf{E}\) back with one derivative. The lab lets you drop charges on a plane and watch the equipotential lines and field lines redraw themselves.

1 · Electric potential energy

Start with something you already know from mechanics. When you lift a book, you do work against gravity, and that work is stored as gravitational potential energy; let go, and it comes back as kinetic energy. This works because gravity is a conservative force: the work it does between two points does not depend on the path taken.

The electrostatic force is conservative too (Coulomb's law has the same \(1/r^2\), centre-pointing form as gravity). So we can define an electric potential energy \(U\) the same way:

\[ \Delta U = U_B - U_A = -W_{A\to B} = -\int_A^B \mathbf{F}\cdot d\boldsymbol{\ell} \] \(W_{A\to B}\) is the work done by the electric force as the charge moves from point \(A\) to point \(B\); \(d\boldsymbol{\ell}\) is a tiny step along the path. If the field does positive work (pushes the charge the way it is going), the potential energy drops, exactly like a falling ball.

For two point charges \(q_1\) and \(q_2\) a distance \(r\) apart, doing that integral with Coulomb's force (and choosing \(U = 0\) when they are infinitely far apart) gives:

\[ U = k\,\frac{q_1 q_2}{r} = \frac{1}{4\pi\varepsilon_0}\frac{q_1 q_2}{r} \] \(k = 8.99\times10^{9}\ \text{N}\cdot\text{m}^2/\text{C}^2\) and \(\varepsilon_0 = 8.85\times10^{-12}\ \text{C}^2/(\text{N}\cdot\text{m}^2)\). Signs matter here, and they are kept: like charges give \(U > 0\) (you had to push them together, the energy is stored, they will fly apart if released); unlike charges give \(U < 0\) (they attract; you would have to supply energy to pull them apart).

For a group of charges, add up one term for every pair (each pair counted once). For three charges:

\[ U = k\left(\frac{q_1q_2}{r_{12}} + \frac{q_1q_3}{r_{13}} + \frac{q_2q_3}{r_{23}}\right) \] This is the total work an outside agent must do to assemble the group, bringing the charges in one at a time from infinity. It is energy stored in the configuration, not in any single charge.

Common mistake

Dropping the signs of the charges “because energy is positive”. Potential energy can be negative, and the sign carries physics: a negative total \(U\) means the system is bound (it would take work to pull it apart). Use the signed values of \(q\) every time, and use \(r\), not \(r^2\) — energy goes as \(1/r\), force as \(1/r^2\).

2 · Electric potential and the volt

Potential energy belongs to a charge and its surroundings. It would be nicer to describe the surroundings alone — the way the field \(\mathbf{E} = \mathbf{F}/q\) describes the force per unit charge. Do the same with energy:

\[ V = \frac{U}{q_0}\qquad\Longleftrightarrow\qquad U = q_0 V,\qquad \Delta U = q_0\,\Delta V \] \(V\) is the electric potential: potential energy per unit of test charge \(q_0\), a property of the location, not of whatever charge you put there. Its unit is the volt, \(1\ \text{V} = 1\ \text{J/C}\). A “voltage” is always a difference in potential, \(\Delta V = V_B - V_A\).

The picture to keep in mind: \(V\) is a landscape. Positive charges roll downhill (toward lower \(V\)), gaining kinetic energy. Negative charges such as electrons do the opposite: since \(U = qV\) with \(q < 0\), a high potential is a low energy for them, so electrons are pulled toward higher \(V\). This is why electrons in a wire flow from the − terminal of a battery toward the + terminal, opposite to the conventional current arrow.

Do not mix up \(V\) and \(U\)

\(V\) (volts, J/C) belongs to a point in space. \(U\) (joules) belongs to a charge sitting at that point. A 9 V battery gives each coulomb 9 J; it gives an electron \(9\ \text{V}\times1.60\times10^{-19}\ \text{C} = 1.44\times10^{-18}\ \text{J}\). Same \(V\), wildly different \(U\).

Only differences of potential are physical. You may choose the zero anywhere (as with heights: sea level or the floor of your room). For isolated charges the usual choice is \(V = 0\) at infinity; in circuits it is the node you call ground.

EE connection

A circuit schematic's ground symbol is nothing more than the choice “\(V = 0\) here”. Every node voltage in SPICE is the potential of that node relative to ground. A multimeter's two probes measure \(\Delta V\) between two points — it cannot measure an absolute \(V\), because there is no such thing. And the unit “V/m” for electric field (equal to N/C) comes straight from this unit.

3 · Potential from the field

Divide \(\Delta U = -\int \mathbf{F}\cdot d\boldsymbol{\ell}\) by \(q_0\) and use \(\mathbf{F} = q_0\mathbf{E}\):

\[ V_B - V_A = -\int_A^B \mathbf{E}\cdot d\boldsymbol{\ell} \] A line integral: walk from \(A\) to \(B\) in small steps \(d\boldsymbol{\ell}\); at each step take the component of \(\mathbf{E}\) along the step times the step length; add them up. The minus sign says: walking along the field, the potential goes down. Because the electrostatic field is conservative, any path from \(A\) to \(B\) gives the same answer — so choose the easiest one.

Uniform field

Between two large parallel plates the field is uniform (Unit 2). Walk a distance \(d\) straight along \(\mathbf{E}\): \(\mathbf{E}\cdot d\boldsymbol{\ell} = E\,d\ell\) at every step, so the integral is just \(Ed\):

\[ \Delta V = -E\,d\quad(\text{moving a distance } d \text{ along } \mathbf{E}),\qquad\qquad E = \frac{|\Delta V|}{d} \] This is why field strength is quoted in volts per metre: 12 V across a 1 mm gap is \(1.2\times10^4\ \text{V/m}\). Moving perpendicular to \(\mathbf{E}\), \(\mathbf{E}\cdot d\boldsymbol{\ell} = 0\) and \(V\) does not change.

A uniform field. The potential falls in equal steps as you walk along \(\mathbf{E}\); the equipotentials (dashed) are planes perpendicular to the field. A proton is pushed down the potential, an electron up it — but both lose potential energy and gain kinetic energy.

Point charge

The field of a point charge \(q\) points radially, \(E = kq/r^2\). Integrate along a radial line from infinity (where we set \(V = 0\)) in to distance \(r\):

\[ V(r) - 0 = -\int_\infty^r \frac{kq}{r'^2}\,dr' = -\left[-\frac{kq}{r'}\right]_\infty^r = \frac{kq}{r} \] \(r'\) is just the integration variable. The result: \(V = kq/r\). It is positive near a positive charge and negative near a negative one, and it falls off as \(1/r\) — more slowly than the field's \(1/r^2\).

Path independence = KVL

If you walk around any closed loop and come back to the start, \(\oint\mathbf{E}\cdot d\boldsymbol{\ell} = 0\) in electrostatics: you return to the same height. In a circuit this is Kirchhoff's voltage law — the voltage rises and drops around any loop sum to zero. (Unit 8 shows where this breaks: a changing magnetic field makes \(\oint\mathbf{E}\cdot d\boldsymbol{\ell}\neq0\), and that is how transformers and generators work.)

4 · Point charges and superposition

\[ V = \sum_i \frac{kq_i}{r_i} \] \(r_i\) is the distance from charge \(q_i\) to the point where you want \(V\). This is a sum of plain numbers, with signs. No components, no angles, no unit vectors. That is the whole reason potential is popular.

Compare: to find \(\mathbf{E}\) from three charges you must find three vectors, break each into \(x\) and \(y\) components, add components and recombine. To find \(V\) you add three numbers. When a problem asks for both, get \(V\) first and think about whether \(\mathbf{E}\) can be read off from it (Section 6).

\(V = 0\) does not mean \(\mathbf{E} = 0\) (and vice versa)

Midway between \(+q\) and \(-q\): \(V = kq/a - kq/a = 0\), but the two fields point the same way and add — the field is strong. Midway between two equal \(+q\) charges: the fields cancel, \(\mathbf{E} = 0\), but \(V = 2kq/a \neq 0\). The field depends on how \(V\) is changing, not on its value — like slope versus height.

The figure below shows the classic case, an electric dipole (\(+q\) and \(-q\) a small distance apart). The dashed curves are equipotentials, the solid curves are field lines. Look at how they always cross at right angles, and how the plane halfway between the charges is the \(V = 0\) equipotential.

Equipotentials and field lines of a dipole, computed from \(V = kq/r_+ - kq/r_-\). Around each charge the equipotentials are nearly circles (close to one charge, the other hardly matters). Every field line crosses every equipotential at 90°, and points from high \(V\) to low \(V\).

5 · Potential of continuous distributions

For a smooth charge distribution (Unit 1's rods, rings and disks), chop it into tiny pieces \(dq\), treat each as a point charge, and add — which, for infinitely many tiny pieces, is an integral:

\[ V = \int \frac{k\,dq}{r},\qquad dq = \lambda\,d\ell\ \ (\text{line}),\quad dq = \sigma\,dA\ \ (\text{surface}),\quad dq = \rho\,dV\ \ (\text{volume}) \] \(r\) is the distance from the piece \(dq\) to the field point. \(\lambda\) (C/m), \(\sigma\) (C/m²) and \(\rho\) (C/m³) are the linear, surface and volume charge densities. Again a scalar integral — no components to cancel by symmetry, because there are no components.

Ring of charge, on its axis

Every piece of the ring is the same distance \(\sqrt{x^2+a^2}\) from \(P\). For the potential, that is all that matters, so the integral is trivial.

\[ V = \int\frac{k\,dq}{\sqrt{x^2+a^2}} = \frac{k}{\sqrt{x^2+a^2}}\int dq = \frac{kQ}{\sqrt{x^2+a^2}} \] \(Q\) is the ring's total charge, \(a\) its radius, \(x\) the distance along the axis from the centre. Checks: at \(x = 0\), \(V = kQ/a\) (every piece is \(a\) away); for \(x \gg a\), \(V \to kQ/x\), a point charge. Compare with Unit 1, where the field of the same ring needed a \(\cos\theta\) factor to kill the sideways components.

Uniformly charged disk, on its axis

Build the disk (radius \(R\), surface density \(\sigma\)) from thin rings of radius \(s\) and width \(ds\). Each ring has charge \(dq = \sigma\,(2\pi s\,ds)\) and, from the ring result, contributes \(k\,dq/\sqrt{x^2+s^2}\):

\[ V = \int_0^R \frac{k\,\sigma\,2\pi s\,ds}{\sqrt{x^2+s^2}} = \pi k\sigma\Big[\,2\sqrt{x^2+s^2}\,\Big]_0^R = \frac{\sigma}{2\varepsilon_0}\left(\sqrt{x^2+R^2} - x\right)\quad (x > 0) \] Using \(k = 1/(4\pi\varepsilon_0)\), so \(2\pi k = 1/(2\varepsilon_0)\). The substitution \(u = x^2+s^2\), \(du = 2s\,ds\) does the integral. Far away (\(x\gg R\)) a binomial expansion gives \(V\approx k(\sigma\pi R^2)/x = kQ/x\) again.

Finite rod, on its own axis

A rod of length \(L\) with uniform \(\lambda\) lies along the \(x\)-axis from \(0\) to \(L\). Find \(V\) at a point a distance \(d\) beyond the end \(x = L\). A piece at position \(x\) is \(L + d - x\) away:

\[ V = \int_0^L \frac{k\lambda\,dx}{L+d-x} = k\lambda\ln\!\left(\frac{L+d}{d}\right) \] Far away, \(\ln(1 + L/d)\approx L/d\), so \(V\approx k\lambda L/d = kQ/d\) — a point charge, as it must be.

Infinite line: when infinity cannot be the zero

For an infinitely long line of charge, \(E = \lambda/(2\pi\varepsilon_0 r)\) (Unit 2). Integrating from a reference distance \(r_0\) to \(r\):

\[ V(r) - V(r_0) = -\int_{r_0}^{r}\frac{\lambda}{2\pi\varepsilon_0 r'}\,dr' = -\frac{\lambda}{2\pi\varepsilon_0}\ln\frac{r}{r_0} \] If you try \(r_0\to\infty\), the logarithm blows up. The charge itself extends to infinity, so “\(V = 0\) at infinity” is not available: pick a finite reference \(r_0\) instead. Only differences are physical anyway. This result returns in Unit 4 as the capacitance of a coaxial cable.

Recipe for \(V\) of a distribution

Sketch, pick a \(dq\), write \(dq\) in terms of a coordinate (\(\lambda\,dx\), \(\sigma\,2\pi s\,ds\), …).
Write the distance \(r\) from \(dq\) to the field point in terms of that coordinate.
Integrate \(k\,dq/r\) over the whole object. Set the limits from the object's extent.
Check a limit: far away it must look like \(kQ/r\).
If the distribution extends to infinity, use \(V = -\int\mathbf{E}\cdot d\boldsymbol{\ell}\) with a finite reference instead.

6 · Field from potential: \(\mathbf{E} = -\nabla V\), and equipotentials

Section 3 went from \(\mathbf{E}\) to \(V\) by integrating. Going back is differentiating. For a tiny step \(dx\) along \(x\), \(dV = -E_x\,dx\), so \(E_x = -\partial V/\partial x\); the same for \(y\) and \(z\):

\[ \mathbf{E} = -\nabla V = -\left(\frac{\partial V}{\partial x}\,\hat{\mathbf{i}} + \frac{\partial V}{\partial y}\,\hat{\mathbf{j}} + \frac{\partial V}{\partial z}\,\hat{\mathbf{k}}\right) \] \(\nabla V\) (“grad \(V\)”) is the gradient: a vector pointing in the direction in which \(V\) increases fastest, with length equal to that rate of increase. \(\partial V/\partial x\) is a partial derivative: differentiate with respect to \(x\) treating \(y\) and \(z\) as constants. The minus sign: the field points downhill, toward decreasing potential.

For a spherically symmetric potential, only the radial derivative survives: \(E_r = -dV/dr\). Check with a point charge: \(-\frac{d}{dr}\left(\frac{kq}{r}\right) = \frac{kq}{r^2}\). Coulomb's law. ✓ Check with the ring: \(E_x = -\frac{d}{dx}\frac{kQ}{\sqrt{x^2+a^2}} = \frac{kQx}{(x^2+a^2)^{3/2}}\), exactly the Unit 1 result, with far less work.

Equipotentials

An equipotential is a surface (or, in a 2-D drawing, a curve) on which \(V\) has the same value everywhere — a contour line on the height map. Three facts follow from \(\mathbf{E} = -\nabla V\):

Field lines cross equipotentials at right angles. Moving along an equipotential, \(dV = 0 = -\mathbf{E}\cdot d\boldsymbol{\ell}\), so \(\mathbf{E}\perp d\boldsymbol{\ell}\).
Crowded equipotentials mean a strong field. If they are drawn at equal \(\Delta V\) steps, \(E\approx|\Delta V|/\Delta s\) where \(\Delta s\) is their spacing.
No work is done moving a charge along an equipotential. \(\Delta U = q\,\Delta V = 0\).

Source	Equipotentials	Field lines
Point charge	Concentric spheres, closer together near the charge	Radial straight lines
Uniform field (parallel plates)	Equally spaced parallel planes	Parallel, equally spaced lines
Dipole	Closed loops around each charge; the bisecting plane is \(V = 0\)	Arcs from \(+q\) to \(-q\)
Any conductor in equilibrium	Its surface is one (Section 7)	Leave the surface perpendicularly

Interactive lab The potential landscape Open it in Interactive Laboratories →

Common mistake

Forgetting the minus sign in \(\mathbf{E} = -\nabla V\), which makes the field point the wrong way. Sanity check every time: near a positive charge, \(V\) is high, so the field must point away from it, toward lower \(V\).

EE connection

Chip and PCB designers think in equipotentials and gradients constantly. A “ground plane” on a circuit board is a big copper equipotential. Inside a MOSFET, the gate voltage shapes the potential landscape in the silicon underneath, and the field \(-\nabla V\) is what pulls carriers into the channel. Field-solver software (for high-voltage insulators, connectors, IC interconnect) computes \(V\) on a grid, then gets \(\mathbf{E}\) by taking the gradient — exactly the order used here.

7 · Potential of conductors

In electrostatic equilibrium the field inside a conductor is zero (Unit 2) — otherwise its free electrons would still be moving. With \(\mathbf{E} = 0\), \(\Delta V = -\int\mathbf{E}\cdot d\boldsymbol{\ell} = 0\) between any two points in the metal. So:

\[ \text{A conductor in equilibrium is an equipotential: the same } V \text{ throughout its volume and over its surface.} \] Consequently the field just outside the surface is perpendicular to it (any sideways component would push charges along the surface). Its magnitude is \(E = \sigma/\varepsilon_0\), from Gauss's law.

A solid or hollow conducting sphere of radius \(R\) carrying charge \(Q\): outside it looks like a point charge, so \(V = kQ/r\) for \(r\ge R\); at and inside the surface \(V = kQ/R\), constant.

Why charge piles up at sharp points

Connect a big sphere (radius \(R_1\)) and a small one (\(R_2\)) with a long thin wire, so they are one conductor at one potential, and far enough apart that each one's charge barely affects the other:

\[ \frac{kq_1}{R_1} = \frac{kq_2}{R_2}\quad\Longrightarrow\quad \frac{q_1}{q_2} = \frac{R_1}{R_2},\qquad \frac{\sigma_1}{\sigma_2} = \frac{q_1/4\pi R_1^2}{q_2/4\pi R_2^2} = \frac{R_2}{R_1} \] The small sphere holds less charge but a higher surface density, and since \(E = \sigma/\varepsilon_0\) at the surface, a stronger field. A sharp point is a sphere of tiny radius: the field there can exceed the breakdown field of air (about \(3\times10^6\ \text{V/m}\)) and spark or glow (corona discharge).

Two spheres joined by a wire share one potential. The small one carries less charge but a stronger surface field \(E = V/R\). This is the lightning rod principle, and why high-voltage hardware is made smooth and rounded.

EE connection

High-voltage engineers fight this effect everywhere: transmission-line hardware has smooth “corona rings”, and high-voltage PCB traces avoid sharp corners because corona discharge wastes power and generates radio noise. On the other hand, electrostatic discharge (ESD) from your fingertip to a pin of a chip can be thousands of volts; ESD protection diodes on every IC input exist because a charged person is a conductor at a high potential looking for somewhere to equalise.

Shielding

Inside an empty cavity in a conductor, \(V\) is constant and \(\mathbf{E} = 0\), no matter what charges sit outside. That is a Faraday cage: shielded cables, the metal can over an RF module, and the screened room used for EMC tests.

8 · Electron-volts and energy conservation

When only electric forces act, the total energy \(K + U = K + qV\) is conserved:

\[ K_A + qV_A = K_B + qV_B\qquad\Longleftrightarrow\qquad \Delta K = -q\,\Delta V \] \(K = \tfrac12 mv^2\) is kinetic energy. A charge that falls through a potential difference gains kinetic energy \(|q\,\Delta V|\). Watch the sign of \(q\): an electron (\(q = -e\)) gains kinetic energy moving to higher \(V\).

Because atomic-scale charges are multiples of \(e = 1.60\times10^{-19}\ \text{C}\), a handy energy unit is the electron-volt: the kinetic energy one electron (or proton) gains by falling through 1 V.

\[ 1\ \text{eV} = (1.60\times10^{-19}\ \text{C})(1\ \text{V}) = 1.60\times10^{-19}\ \text{J} \] The trick: a particle of charge \(ze\) accelerated through \(\Delta V\) volts gains exactly \(z\,\Delta V\) eV. A proton through 2000 V gains 2000 eV = 2 keV; an alpha particle (\(z = 2\)) through the same voltage gains 4 keV. Convert to joules only when you need a speed.

Common mistake

The electron-volt is a unit of energy, not of potential. “The electron has 500 V” is wrong; “the electron was accelerated through 500 V and has 500 eV” is right. Also: use the particle's own mass (\(m_e = 9.11\times10^{-31}\) kg, \(m_p = 1.67\times10^{-27}\) kg) when you convert to speed.

EE connection

Semiconductor physics is written in eV. Silicon's band gap is about 1.12 eV; a silicon pn junction's built-in barrier is roughly 0.7 V, so an electron needs about 0.7 eV to climb it, compared with a thermal energy of only \(k_BT\approx0.026\) eV at room temperature (here \(k_B\) is Boltzmann's constant, not Coulomb's). Forward bias lowers that barrier, and the diode turns on once most of it is gone — that is why silicon diodes conduct at about 0.6–0.7 V. Vacuum tubes, X-ray tubes and electron microscopes are all “accelerate through \(\Delta V\), gain \(e\,\Delta V\)” machines.

9 · Worked examples

1. An electron gun

An electron starts from rest and is accelerated through a potential difference of 500 V. Find its kinetic energy (eV and J) and its final speed.

Charge \(q = -e\); it moves to a point 500 V higher, so \(\Delta U = q\,\Delta V = (-e)(+500\ \text{V})\), a decrease of 500 eV. Energy conservation:

\[ K = 500\ \text{eV} = 500\times1.60\times10^{-19}\ \text{J} = 8.00\times10^{-17}\ \text{J} \] \[ v = \sqrt{\frac{2K}{m_e}} = \sqrt{\frac{2(8.00\times10^{-17})}{9.11\times10^{-31}}} = \sqrt{1.756\times10^{14}}\ \text{m/s} = \mathbf{1.33\times10^{7}\ m/s} \]

About 4.4% of the speed of light, so ignoring relativity is fine. (Above roughly 10 kV for electrons, relativistic corrections start to matter.)

2. Potential of two point charges, and the work to bring in a third

\(q_1 = +4.0\ \text{nC}\) sits at the origin and \(q_2 = -2.0\ \text{nC}\) at \((0.30\ \text{m}, 0)\). (a) Find \(V\) at \(P = (0, 0.40\ \text{m})\). (b) How much work must you do to bring \(q_3 = +1.0\ \text{nC}\) from infinity to \(P\)?

Distances: \(r_1 = 0.40\) m; \(r_2 = \sqrt{0.30^2 + 0.40^2} = 0.50\) m.

\[ V_P = k\left(\frac{q_1}{r_1} + \frac{q_2}{r_2}\right) = (8.99\times10^9)\left(\frac{4.0\times10^{-9}}{0.40} - \frac{2.0\times10^{-9}}{0.50}\right) = (8.99\times10^9)(6.0\times10^{-9}) = \mathbf{53.9\ V} \] \[ W = q_3(V_P - V_\infty) = (1.0\times10^{-9})(53.9) = \mathbf{5.39\times10^{-8}\ J} \]

Positive work: \(P\) is at higher potential, so you push the positive charge “uphill”. Note the negative charge entered with its sign — it lowers \(V\).

3. Field from a given potential

In a region, \(V(x,y,z) = 3x^2y - 2yz^2 + 5z\) (volts, with coordinates in metres). Find \(\mathbf{E}\) and \(|\mathbf{E}|\) at \((1, 2, -1)\).

\[ \frac{\partial V}{\partial x} = 6xy = 12,\qquad \frac{\partial V}{\partial y} = 3x^2 - 2z^2 = 3 - 2 = 1,\qquad \frac{\partial V}{\partial z} = -4yz + 5 = 8 + 5 = 13 \] \[ \mathbf{E} = -\nabla V = \mathbf{(-12\,\hat{i} - 1\,\hat{j} - 13\,\hat{k})\ V/m},\qquad |\mathbf{E}| = \sqrt{144 + 1 + 169} = \sqrt{314} = \mathbf{17.7\ V/m} \]

Careful with \(\partial V/\partial z\): \(-2yz^2\) gives \(-4yz = -4(2)(-1) = +8\). Sign slips in the substitution are the usual way to lose marks here.

4. A charged ring: potential, field, and a proton launched from its centre

A ring of radius \(a = 0.10\) m carries \(Q = +5.0\) nC. (a) Find \(V\) and \(E\) on the axis at \(x = 0.24\) m. (b) A proton is released (essentially) from rest at the centre and pushed off along the axis. What speed does it reach far away?

(a) \(\sqrt{x^2+a^2} = \sqrt{0.0576 + 0.0100} = \sqrt{0.0676} = 0.26\) m, and \(kQ = (8.99\times10^9)(5.0\times10^{-9}) = 44.95\ \text{V}\cdot\text{m}\).

\[ V = \frac{kQ}{\sqrt{x^2+a^2}} = \frac{44.95}{0.26} = \mathbf{173\ V},\qquad E_x = \frac{kQx}{(x^2+a^2)^{3/2}} = \frac{44.95\times0.24}{0.26^3} = \frac{10.79}{0.01758} = \mathbf{614\ V/m} \]

(b) At the centre \(V = kQ/a = 44.95/0.10 = 449.5\) V (and \(E = 0\) by symmetry, so the proton needs a tiny nudge). Far away \(V = 0\). The proton falls through 449.5 V and gains 449.5 eV:

\[ v = \sqrt{\frac{2e\,\Delta V}{m_p}} = \sqrt{\frac{2(1.60\times10^{-19})(449.5)}{1.67\times10^{-27}}} = \sqrt{8.61\times10^{10}} = \mathbf{2.93\times10^{5}\ m/s} \]

Energy conservation needed no knowledge of how the force varied along the way — only the start and end potentials. That is the power of \(V\).

5. Two spheres joined by a wire

Conducting spheres of radius \(R_1 = 0.10\) m and \(R_2 = 0.050\) m, far apart, are joined by a thin wire and share a total charge of 30 nC. Find each charge, the common potential, and each surface field.

\[ \frac{q_1}{q_2} = \frac{R_1}{R_2} = 2,\quad q_1 + q_2 = 30\ \text{nC}\ \Rightarrow\ \mathbf{q_1 = 20\ nC,\ q_2 = 10\ nC} \] \[ V = \frac{kq_1}{R_1} = \frac{(8.99\times10^9)(20\times10^{-9})}{0.10} = \mathbf{1.80\ kV}\quad\left(\text{check: } \frac{kq_2}{R_2} = \frac{89.9}{0.050} = 1798\ \text{V}\ \checkmark\right) \] \[ E_1 = \frac{V}{R_1} = \mathbf{1.80\times10^{4}\ V/m},\qquad E_2 = \frac{V}{R_2} = \mathbf{3.60\times10^{4}\ V/m} \]

Half the radius, twice the surface field. (\(E = kq/R^2 = V/R\) at the surface of a sphere.)

6. Closest approach: Rutherford scattering

An alpha particle (charge \(+2e\)) with kinetic energy 5.0 MeV heads straight at a gold nucleus (charge \(+79e\)), which stays fixed. How close does it get?

At closest approach the alpha is momentarily at rest: all kinetic energy has become potential energy.

\[ K = \frac{k(2e)(79e)}{r_{\min}}\quad\Rightarrow\quad r_{\min} = \frac{k(2)(79)e^2}{K} \]

Use the eV trick: \(K = 5.0\times10^6\ \text{eV} = (5.0\times10^6\ \text{V})\,e\), so one factor of \(e\) cancels:

\[ r_{\min} = \frac{(8.99\times10^9)(158)(1.60\times10^{-19})}{5.0\times10^{6}} = \mathbf{4.5\times10^{-14}\ m} \]

About 45 fm — a few times the gold nucleus's actual radius (about 7 fm). This is how Rutherford concluded that the positive charge of an atom is packed into a tiny nucleus.

Checklist

I can explain \(V\) as potential energy per unit charge, and I never confuse V (volts) with U (joules).
I keep signs: \(U = kq_1q_2/r\) and \(V = kq/r\) both carry the sign of the charges.
I can find \(\Delta V = -\int\mathbf{E}\cdot d\boldsymbol{\ell}\), including \(\Delta V = -Ed\) in a uniform field.
I add potentials as numbers, and I know \(V = 0\) and \(\mathbf{E} = 0\) are different statements.
I can set up \(V = \int k\,dq/r\) for a ring, a disk and a rod, and check the far-field limit.
I can get \(\mathbf{E} = -\nabla V\) from a formula for \(V\), and I sketch equipotentials perpendicular to field lines.
I know a conductor is an equipotential and that sharp points have the strongest fields.
I use energy conservation and electron-volts for speeds and closest approaches.