Practice Exam 4

Capacitance & Dielectrics

Questions10

Time90 min

Points100

Formula sheetProvided

Use \(k = 8.99\times10^{9}\ \text{N}\cdot\text{m}^2/\text{C}^2\), \(\varepsilon_0 = 8.85\times10^{-12}\ \text{F/m}\), \(e = 1.60\times10^{-19}\ \text{C}\). Breakdown field of air: \(3.0\times10^{6}\) V/m.

1What C depends on (conceptual)8 pts

A student says: “Since \(C = Q/V\), doubling the voltage on a capacitor halves its capacitance.” Explain what is wrong, state what actually happens to \(Q\) and to the stored energy when \(V\) is doubled, and list what \(C\) does depend on for a parallel-plate capacitor.

Solution

\(Q\) and \(V\) are proportional: doubling \(V\) doubles the field everywhere, which requires double the charge. The ratio \(C\) is unchanged. So \(Q\to2Q\), and

\[ U = \tfrac12CV^2\ \to\ \tfrac12C(2V)^2 = \mathbf{4U} \]

For parallel plates \(C = \kappa\varepsilon_0A/d\): it depends only on plate area \(A\), gap \(d\) and the dielectric \(\kappa\). (Doubling \(V\) does eventually matter if it pushes \(E = V/d\) past the dielectric strength — but then the capacitor breaks down rather than changing \(C\).)

2Parallel-plate design10 pts

You need a 100 pF air-gap capacitor with plates 0.50 mm apart. (a) What plate area is needed? (b) What is the maximum voltage before the air breaks down?

Solution

\[ \text{(a)}\quad A = \frac{Cd}{\varepsilon_0} = \frac{(100\times10^{-12})(0.50\times10^{-3})}{8.85\times10^{-12}} = \mathbf{5.65\times10^{-3}\ m^2 \approx 56.5\ cm^2} \] \[ \text{(b)}\quad V_{\max} = E_{\max}d = (3.0\times10^{6})(0.50\times10^{-3}) = \mathbf{1.5\ kV} \]

A square about 7.5 cm on a side for only 100 pF — which is why practical capacitors use dielectrics and rolled or stacked layers.

3Derivation: spherical capacitor12 pts

A conducting sphere of radius \(a\) sits at the centre of a thin conducting shell of radius \(b\). (a) Use Gauss's law and the potential to derive \(C\). (b) Evaluate for \(a = 5.0\) cm, \(b = 6.0\) cm. (c) Show that your result reduces to the isolated-sphere result as \(b\to\infty\) and to the parallel-plate result when \(b - a\ll a\).

Solution

(a) Put \(+Q\) on the inner sphere and \(-Q\) on the shell. A Gaussian sphere of radius \(a<r<b\) encloses \(Q\): \(E\,4\pi r^2 = Q/\varepsilon_0\), so \(E = kQ/r^2\), radially outward.

\[ V = V_a - V_b = \int_a^b\frac{kQ}{r^2}\,dr = kQ\left(\frac1a - \frac1b\right) = \frac{kQ(b-a)}{ab}\quad\Rightarrow\quad C = \frac{Q}{V} = \mathbf{4\pi\varepsilon_0\frac{ab}{b-a}} \] \[ \text{(b)}\quad C = \frac{1}{k}\cdot\frac{(0.050)(0.060)}{0.010} = \frac{0.30\ \text{m}}{8.99\times10^{9}\ \text{m/F}} = \mathbf{33.4\ pF} \]

(c) As \(b\to\infty\), \(ab/(b-a)\to a\), so \(C\to4\pi\varepsilon_0a\) ✓ (isolated sphere). If \(b - a = d\ll a\), then \(ab\approx a^2\) and \(C\approx\varepsilon_0(4\pi a^2)/d = \varepsilon_0A/d\) ✓ (parallel plates with the sphere's area).

4Coaxial cable10 pts

A coax cable has inner radius 0.50 mm, outer radius 2.0 mm and Teflon insulation (\(\kappa = 2.1\)). Find its capacitance per metre and the total capacitance of a 25 m run.

Solution

\[ \frac{C}{L} = \frac{2\pi\kappa\varepsilon_0}{\ln(b/a)} = \frac{2\pi(2.1)(8.85\times10^{-12})}{\ln 4} = \frac{1.168\times10^{-10}}{1.386} = \mathbf{84.2\ pF/m} \] \[ C = (84.2\ \text{pF/m})(25\ \text{m}) = \mathbf{2.11\ nF} \]

Only the ratio \(b/a = 4\) mattered, not the absolute size. A 2 nF load is significant for a signal source: it has to charge and discharge that capacitance on every edge, which is why long cables slow down signals unless they are driven and terminated as transmission lines.

5Capacitor network10 pts

\(C_1 = 2.0\ \mu\text{F}\) and \(C_2 = 3.0\ \mu\text{F}\) are in parallel; that pair is in series with \(C_3 = 5.0\ \mu\text{F}\). A 10 V source is across the whole network. Find \(C_{\text{eq}}\), the charge on each capacitor, and the total energy. Verify the energy by adding the three individual energies.

Solution

\[ C_{12} = 2.0 + 3.0 = 5.0\ \mu\text{F},\qquad C_{\text{eq}} = \frac{(5.0)(5.0)}{5.0 + 5.0} = \mathbf{2.5\ \mu F} \] \[ Q_3 = Q_{12} = C_{\text{eq}}V = (2.5)(10) = \mathbf{25\ \mu C},\qquad V_3 = \frac{25}{5.0} = 5.0\ \text{V},\qquad V_{12} = 10 - 5.0 = 5.0\ \text{V} \] \[ Q_1 = (2.0)(5.0) = \mathbf{10\ \mu C},\qquad Q_2 = (3.0)(5.0) = \mathbf{15\ \mu C} \] \[ U = \tfrac12(2.5\times10^{-6})(10)^2 = \mathbf{125\ \mu J} \]

Check: \(U_1 + U_2 + U_3 = \tfrac12(2.0)(5.0)^2 + \tfrac12(3.0)(5.0)^2 + \tfrac12(5.0)(5.0)^2 = 25 + 37.5 + 62.5 = 125\ \mu\text{J}\) ✓. Also \(Q_1 + Q_2 = Q_3\) ✓ (the node between them is isolated).

6Energy density10 pts

An air parallel-plate capacitor with \(A = 100\ \text{cm}^2\) and \(d = 2.0\) mm is held at 4.0 kV. (a) Find the field and the energy density between the plates. (b) Find the stored energy two ways: from \(u\) and from \(\tfrac12CV^2\). (c) Is the air close to breakdown?

Solution

\[ \text{(a)}\quad E = \frac{V}{d} = \frac{4000}{2.0\times10^{-3}} = \mathbf{2.0\times10^{6}\ V/m},\qquad u = \tfrac12\varepsilon_0E^2 = \tfrac12(8.85\times10^{-12})(4.0\times10^{12}) = \mathbf{17.7\ J/m^3} \] \[ \text{(b)}\quad U = u(Ad) = (17.7)(0.010)(2.0\times10^{-3}) = \mathbf{3.54\times10^{-4}\ J} \] \[ C = \frac{\varepsilon_0A}{d} = \frac{(8.85\times10^{-12})(0.010)}{2.0\times10^{-3}} = 44.3\ \text{pF},\qquad \tfrac12CV^2 = \tfrac12(4.425\times10^{-11})(1.6\times10^{7}) = 3.54\times10^{-4}\ \text{J}\ \checkmark \]

(c) \(2.0\) MV/m is two thirds of air's 3.0 MV/m breakdown field. It works on paper, but with humidity, dust or a sharp edge on a plate, it is uncomfortably close — a designer would want a larger gap or a solid dielectric.

7Partially filled capacitor10 pts

An air parallel-plate capacitor has \(C_0 = 10\) pF. A slab with \(\kappa = 3.0\) is inserted in two different ways: (a) it fills the full gap but covers only half the plate area; (b) it covers the whole area but fills only half the gap (thickness \(d/2\)). Find \(C\) in each case.

Solution

(a) Side by side: two capacitors of area \(A/2\) sharing the same plates, i.e. the same \(V\) — parallel. Each half has \(C_0/2\) of vacuum capacitance:

\[ C = \frac{C_0}{2} + \kappa\frac{C_0}{2} = \frac{(1+\kappa)}{2}C_0 = \frac{4.0}{2}(10) = \mathbf{20\ pF} \]

(b) Stacked: an air layer and a dielectric layer, each of thickness \(d/2\), carry the same \(Q\) — series. Halving the gap doubles the capacitance, so the layers are \(2C_0 = 20\) pF and \(2\kappa C_0 = 60\) pF:

\[ C = \frac{(20)(60)}{20 + 60} = \mathbf{15\ pF}\qquad\left(\text{in general } \frac{2\kappa}{1+\kappa}C_0\right) \]

The same amount of dielectric helps more side by side. In (b) the air layer, with its weaker \(\varepsilon\), takes most of the voltage and limits the total.

8Connected vs isolated (conceptual)10 pts

A parallel-plate capacitor is charged by a battery. Its plate separation is then doubled. Fill in what happens to \(C\), \(Q\), \(V\), \(E\) and \(U\) (a) if the battery is still connected; (b) if it was disconnected first. In (b), where does the change in energy come from?

Solution

\(C = \varepsilon_0A/d\) halves in both cases.

(a) Connected, \(V\) fixed: \(Q = CV\) halves; \(V\) same; \(E = V/d\) halves; \(U = \tfrac12CV^2\) halves. (Charge flows back into the battery.)

(b) Isolated, \(Q\) fixed: \(Q\) same; \(V = Q/C\) doubles; \(E = \sigma/\varepsilon_0\) same; \(U = Q^2/2C\) doubles.

In (b) the extra energy is the work you did pulling the plates apart against their attraction (the plates attract with a force \(Q^2/(2\varepsilon_0A)\), constant at fixed \(Q\)). The extra energy lives in the newly created volume of field: same \(u\), twice the volume.

9EE application: reading a DRAM cell12 pts

A DRAM cell is a 25 fF capacitor storing a “1” at 1.0 V or a “0” at 0 V. To read it, it is connected to a bit line with capacitance 250 fF that has been precharged to 0.50 V. (a) How many electrons' worth of charge does a stored “1” hold? (b) Find the bit-line voltage after connection for a stored “1” and for a “0”. (c) Why must a sense amplifier then rewrite the cell?

Solution

\[ \text{(a)}\quad Q = CV = (25\times10^{-15})(1.0) = 25\ \text{fC},\qquad N = \frac{25\times10^{-15}}{1.60\times10^{-19}} = \mathbf{1.6\times10^{5}\ electrons} \]

(b) Connecting puts the two capacitors in parallel; total charge is conserved:

\[ V_f = \frac{C_{\text{cell}}V_{\text{cell}} + C_{\text{BL}}V_{\text{BL}}}{C_{\text{cell}} + C_{\text{BL}}} \] \[ \text{"1"}:\ V_f = \frac{(25)(1.0) + (250)(0.50)}{275} = \mathbf{0.545\ V}\ (+45\ \text{mV}),\qquad \text{"0"}:\ V_f = \frac{0 + 125}{275} = \mathbf{0.455\ V}\ (-45\ \text{mV}) \]

A 1-volt bit produces only a ±45 mV wiggle on the bit line, because the bit line's capacitance is ten times the cell's. A sense amplifier compares it to 0.50 V to decide.

(c) Charge sharing destroyed the stored value: the cell now sits at about 0.5 V either way. The read is destructive, so the amplified result is written back. (Cells also leak, which is why DRAM needs periodic refresh.)

10Maximum energy density8 pts

Compare the maximum energy density that can be stored in the field of (a) air and (b) polystyrene (\(\kappa = 2.56\), dielectric strength 24 MV/m), each just below breakdown. Use \(u = \tfrac12\kappa\varepsilon_0E^2\).

Solution

\[ \text{(a)}\quad u_{\text{air}} = \tfrac12(1)(8.85\times10^{-12})(3.0\times10^{6})^2 = \mathbf{39.8\ J/m^3} \] \[ \text{(b)}\quad u_{\text{PS}} = \tfrac12(2.56)(8.85\times10^{-12})(24\times10^{6})^2 = \mathbf{6.5\times10^{3}\ J/m^3} \]

About 160 times more: a factor 2.56 from \(\kappa\) and a factor \((24/3)^2 = 64\) from the higher allowed field. The field-strength term dominates — when choosing a dielectric for energy storage, breakdown strength matters more than \(\kappa\).

Scoring

Score	Where you are
90–100	Solid. On to current and DC circuits, where capacitors meet resistors.
75–89	Mostly right; check series/parallel slips (Q5, Q7) and the “which is fixed” decision (Q8). Review Section 4 and Section 8.
60–74	Derivations (Q3, Q4) or energy (Q6, Q10) are shaky. Re-read Section 3 and Section 5, and redo worked examples 1–2.
< 60	Restart from Section 1 and Section 2; if Gauss's law or \(V = -\int\mathbf{E}\cdot d\boldsymbol{\ell}\) feel unsure, revisit Unit 3, Section 3 first. Then try the dielectric lab.

Where the points usually go

Adding capacitors in series as if they were resistors (Q5, Q7); using \(U = \tfrac12CV^2\) when \(Q\) is the fixed quantity, which gives the wrong trend (Q8); forgetting to convert cm² to m² or mm to m (Q2, Q6); and in charge-sharing problems (Q9), assuming voltage rather than charge is conserved.