Practice Exam 3

Electric Potential

Questions10

Time90 min

Points100

CalculatorAllowed

Use \(k = 8.99\times10^{9}\ \text{N}\cdot\text{m}^2/\text{C}^2\), \(\varepsilon_0 = 8.85\times10^{-12}\ \text{C}^2/(\text{N}\cdot\text{m}^2)\), \(e = 1.60\times10^{-19}\ \text{C}\), \(m_e = 9.11\times10^{-31}\ \text{kg}\), \(m_p = 1.67\times10^{-27}\ \text{kg}\).

1Potential vs field (conceptual)8 pts

Two charges sit a distance \(2a\) apart. Point \(M\) is the midpoint. For each case, say whether \(V_M\) is zero and whether \(\mathbf{E}_M\) is zero, and explain in one sentence: (a) \(+q\) and \(-q\); (b) \(+q\) and \(+q\). (Take \(V = 0\) at infinity.)

Solution

\[ \text{(a)}\quad V_M = \frac{kq}{a} + \frac{k(-q)}{a} = 0,\qquad E_M = \frac{kq}{a^2} + \frac{kq}{a^2} = \frac{2kq}{a^2}\ \text{(toward } -q) \] \[ \text{(b)}\quad V_M = \frac{2kq}{a}\neq0,\qquad E_M = \frac{kq}{a^2} - \frac{kq}{a^2} = 0 \]

(a) \(V = 0\), \(\mathbf{E}\neq0\): potentials add as signed numbers and cancel, but both field vectors point from \(+q\) toward \(-q\) and add. (b) \(V\neq0\), \(\mathbf{E} = 0\): the field vectors are opposite and cancel, but the two positive potentials add. The field depends on the slope of \(V\), not its value.

2Uniform field10 pts

A uniform field \(\mathbf{E} = (2.0\times10^{3}\ \text{V/m})\,\hat{\mathbf{i}}\) fills a region. (a) Find \(V(0.30\ \text{m}) - V(0)\) along the \(x\)-axis. (b) Find the change in potential energy of a proton and of an electron moved from \(x = 0\) to \(x = 0.30\) m, in eV and J. (c) An electron is released from rest at \(x = 0.30\) m. Which way does it move, and what is its kinetic energy when it reaches \(x = 0\)?

Solution

\[ \text{(a)}\quad \Delta V = -E\,\Delta x = -(2.0\times10^{3})(0.30) = \mathbf{-600\ V} \] \[ \text{(b)}\quad \Delta U_p = (+e)(-600\ \text{V}) = \mathbf{-600\ eV = -9.6\times10^{-17}\ J},\qquad \Delta U_e = (-e)(-600\ \text{V}) = \mathbf{+600\ eV = +9.6\times10^{-17}\ J} \]

The proton moves with the field (the way the force pushes it), so it loses potential energy; the electron is dragged against its force, so it gains.

(c) The force on an electron is \(-e\mathbf{E}\), in the \(-x\) direction, so it moves toward \(x = 0\), i.e. toward higher potential. \(\Delta K = -q\,\Delta V = -(-e)(+600\ \text{V}) = \mathbf{600\ eV = 9.6\times10^{-17}\ J}\).

3Superposition & system energy10 pts

Charges \(+2.0\) nC, \(+2.0\) nC and \(-1.0\) nC sit at the corners of an equilateral triangle of side \(s = 0.20\) m. (a) Find \(V\) at the triangle's centre. (b) Find the electrostatic potential energy of the system. (c) Interpret the sign of your answer to (b).

Solution

(a) The centre is \(r = s/\sqrt3 = 0.1155\) m from each corner, so

\[ V = \frac{k(q_1+q_2+q_3)}{r} = \frac{(8.99\times10^{9})(3.0\times10^{-9})}{0.1155} = \frac{26.97}{0.1155} = \mathbf{234\ V} \]

(b) One term per pair, all separated by \(s\):

\[ U = \frac{k}{s}\left(q_1q_2 + q_1q_3 + q_2q_3\right) = \frac{8.99\times10^{9}}{0.20}\left[(2)(2) + (2)(-1) + (2)(-1)\right]\times10^{-18} = \mathbf{0} \]

(c) Zero total work is needed to assemble this arrangement from infinitely far apart: the positive work spent pushing the two \(+2\) nC charges together is exactly repaid by the attraction of the \(-1\) nC charge. Note that \(V\) at the centre is nonzero even though \(U\) is zero — they answer different questions.

4\(V\) from \(\mathbf{E}\) by integration10 pts

An insulating sphere of radius \(R\) carries charge \(Q\) spread uniformly through its volume. Its field is \(E = kQr/R^3\) inside and \(kQ/r^2\) outside (Unit 2). (a) Derive \(V(r)\) for \(r\le R\), with \(V = 0\) at infinity. (b) Evaluate \(V\) at the surface and at the centre for \(Q = 4.0\) nC, \(R = 5.0\) cm.

Solution

Outside, the sphere acts like a point charge, so \(V(R) = kQ/R\). Then go inward from \(R\) to \(r\):

\[ V(r) = V(R) - \int_R^r \frac{kQr'}{R^3}\,dr' = \frac{kQ}{R} - \frac{kQ}{2R^3}\left(r^2 - R^2\right) = \mathbf{\frac{kQ}{2R}\left(3 - \frac{r^2}{R^2}\right)} \]

Check: at \(r = R\) this gives \(kQ/R\) ✓ (continuous).

\[ \text{(b)}\quad V(R) = \frac{(8.99\times10^9)(4.0\times10^{-9})}{0.050} = \mathbf{719\ V},\qquad V(0) = \frac{3}{2}\cdot719.2 = \mathbf{1.08\ kV} \]

The centre is at the highest potential: the field points outward everywhere, so walking inward is walking against the field, and \(V\) rises.

5Charged disk on its axis10 pts

A disk of radius \(R = 0.10\) m carries uniform surface charge \(\sigma = 5.0\times10^{-8}\ \text{C/m}^2\). Its axial potential is \(V = \dfrac{\sigma}{2\varepsilon_0}\left(\sqrt{x^2+R^2} - x\right)\). (a) Find \(V\) at \(x = 0.10\) m. (b) Differentiate to find \(E_x(x)\), and evaluate it at the same point.

Solution

\[ \frac{\sigma}{2\varepsilon_0} = \frac{5.0\times10^{-8}}{2(8.85\times10^{-12})} = 2825\ \text{V/m} \] \[ \text{(a)}\quad V = 2825\left(\sqrt{0.0200} - 0.10\right) = 2825(0.04142) = \mathbf{117\ V} \] \[ \text{(b)}\quad E_x = -\frac{dV}{dx} = \frac{\sigma}{2\varepsilon_0}\left(1 - \frac{x}{\sqrt{x^2+R^2}}\right) = 2825\,(1 - 0.7071) = \mathbf{827\ V/m} \]

As \(R\to\infty\), \(E_x\to\sigma/2\varepsilon_0\): the infinite sheet of Unit 2, recovered from one derivative.

6Gradient & force10 pts

In a plane, \(V(x,y) = 4x^2y - 3y^3\) (V, metres). At \((2, 1)\): (a) find \(\mathbf{E}\) and \(|\mathbf{E}|\); (b) find the force on a \(+3.0\ \mu\text{C}\) charge there; (c) give a unit vector along which \(V\) does not change.

Solution

\[ \frac{\partial V}{\partial x} = 8xy = 16,\qquad \frac{\partial V}{\partial y} = 4x^2 - 9y^2 = 16 - 9 = 7 \] \[ \text{(a)}\quad \mathbf{E} = -\nabla V = \mathbf{(-16\,\hat{i} - 7\,\hat{j})\ V/m},\qquad |\mathbf{E}| = \sqrt{256 + 49} = \sqrt{305} = \mathbf{17.5\ V/m} \] \[ \text{(b)}\quad \mathbf{F} = q\mathbf{E} = (3.0\times10^{-6})(-16, -7) = \mathbf{(-48\,\hat{i} - 21\,\hat{j})\ \mu N},\quad |\mathbf{F}| = \mathbf{52.4\ \mu N} \]

(c) Along an equipotential, i.e. perpendicular to \(\nabla V = (16, 7)\): \(\mathbf{u} = \mathbf{(7\,\hat{i} - 16\,\hat{j})/\sqrt{305}}\) (or its negative). Check: \((16)(7) + (7)(-16) = 0\) ✓.

7Energy conservation10 pts

A proton moving at \(4.0\times10^{5}\) m/s heads toward a region whose potential is 1.0 kV higher than where it starts. (a) Does it get there? If not, how much potential does it climb before stopping? (b) What minimum initial speed would it need?

Solution

\[ K = \tfrac12 m_pv^2 = \tfrac12(1.67\times10^{-27})(4.0\times10^{5})^2 = 1.336\times10^{-16}\ \text{J} = \frac{1.336\times10^{-16}}{1.60\times10^{-19}}\ \text{eV} = 835\ \text{eV} \]

(a) Climbing 1.0 kV costs a proton 1000 eV; it has only 835 eV. It does not make it: it stops after climbing 835 V, then is pushed back.

\[ \text{(b)}\quad v_{\min} = \sqrt{\frac{2e\,\Delta V}{m_p}} = \sqrt{\frac{2(1.60\times10^{-19})(1000)}{1.67\times10^{-27}}} = \sqrt{1.916\times10^{11}} = \mathbf{4.38\times10^{5}\ m/s} \]

8Conductors & breakdown10 pts

The dome of a Van de Graaff generator is a conducting sphere of radius 15 cm in air, which breaks down at about \(3.0\times10^{6}\) V/m. (a) What is the highest potential the dome can reach, and what charge does it then hold? (b) Why would a dome with a small sharp bump on it reach a much lower potential?

Solution

At the surface of a sphere, \(E = kQ/R^2\) and \(V = kQ/R\), so \(V = ER\).

\[ V_{\max} = E_{\max}R = (3.0\times10^{6})(0.15) = \mathbf{4.5\times10^{5}\ V},\qquad Q = \frac{V_{\max}R}{k} = \frac{(4.5\times10^{5})(0.15)}{8.99\times10^{9}} = \mathbf{7.5\ \mu C} \]

(b) The whole conductor is one equipotential, but the bump has a small radius of curvature, and \(E\approx V/r\) for a small local radius \(r\). The field at the bump reaches breakdown at a much lower \(V\), and the dome leaks charge there by corona discharge. That is why high-voltage electrodes are large and polished.

9EE application: the pn junction barrier12 pts

In a silicon diode at equilibrium, the n side sits at a potential 0.70 V above the p side (the built-in potential), across a depletion region 0.50 μm wide. (a) Estimate the average field in the depletion region and give its direction. (b) How much energy (eV and J) does a conduction electron need to cross from the n side to the p side? (c) Compare with the typical thermal energy \(k_BT\approx0.026\) eV at room temperature. (d) A forward bias of 0.60 V reduces the potential difference across the junction to 0.10 V. What is the new barrier?

Solution

\[ \text{(a)}\quad E_{\text{avg}} = \frac{\Delta V}{d} = \frac{0.70}{0.50\times10^{-6}} = \mathbf{1.4\times10^{6}\ V/m},\ \text{pointing from n to p (high } V \text{ to low } V) \]

(b) The electron moves from high \(V\) to low \(V\); with \(q = -e\), \(\Delta U = (-e)(-0.70\ \text{V}) = +0.70\) eV. It needs at least 0.70 eV = \(1.12\times10^{-19}\) J of kinetic energy.

(c) \(0.70/0.026\approx27\): the barrier is about 27 times the typical thermal energy, so only a tiny fraction of electrons can cross — the diode barely conducts.

(d) The barrier becomes 0.10 eV, only about \(4k_BT\). Many more electrons now have enough energy, and current flows. (The number that can cross grows exponentially as the barrier drops, which is why diode current rises so steeply around 0.6–0.7 V.) Note that 1.4 MV/m is close to half the breakdown field of air — but it exists across less than a micrometre of solid silicon.

10Derivation: finite rod10 pts

A rod of length \(L\) with uniform charge density \(\lambda\) (total \(Q = \lambda L\)) lies on the \(x\)-axis from \(0\) to \(L\). (a) Derive \(V\) at a point on the axis a distance \(d\) beyond \(x = L\). (b) Show that it reduces to a point charge for \(d\gg L\). (c) Differentiate to find the field there.

Solution

(a) A piece \(dq = \lambda\,dx\) at position \(x\) is a distance \(L + d - x\) from the point:

\[ V = \int_0^L\frac{k\lambda\,dx}{L+d-x} = k\lambda\Big[-\ln(L+d-x)\Big]_0^L = \mathbf{k\lambda\ln\frac{L+d}{d}} \] \[ \text{(b)}\quad \ln\!\left(1 + \frac{L}{d}\right)\approx\frac{L}{d}\ \ (d\gg L)\quad\Rightarrow\quad V\approx\frac{k\lambda L}{d} = \frac{kQ}{d}\ \checkmark \] \[ \text{(c)}\quad E_x = -\frac{dV}{dd} = -k\lambda\left(\frac{1}{L+d} - \frac{1}{d}\right) = \mathbf{\frac{kQ}{d(d+L)}}\ \text{(pointing away from the rod for } Q > 0) \]

Far away \(d(d+L)\to d^2\), so \(E\to kQ/d^2\) ✓. Here \(d\) is the variable distance, so \(-dV/dd\) is the derivative with respect to the coordinate along the axis.

Scoring

Score	Where you are
90–100	Solid. On to capacitance, where everything here gets used.
75–89	Concepts fine; sign slips. Re-check the sign of \(q\) in \(\Delta U = q\,\Delta V\) (Q2, Q9) and the minus in \(\mathbf{E} = -\nabla V\) (Q5, Q6). See Section 6.
60–74	Integration set-ups (Q4, Q10) or energy conservation (Q7) are shaky. Re-read Section 5 and Section 8, and redo the worked examples.
< 60	Start again from Section 2 and Section 3: \(V\) versus \(U\), and \(\Delta V = -\int\mathbf{E}\cdot d\boldsymbol{\ell}\), must be automatic first. Use the lab to build the picture.

Where the points usually go

Dropping the sign of a negative charge in a superposition sum (Q3); writing \(\Delta U = e\,\Delta V\) for an electron instead of \(-e\,\Delta V\) (Q2, Q9); using \(1/r^2\) in a potential or \(1/r\) in a field; and confusing “\(V = 0\)” with “\(\mathbf{E} = 0\)” (Q1). On conductors (Q8), remember the whole object is one potential — the field is what varies over its surface.