Electricity and Magnetism 203-SN2-RE • CEGEP, Montreal

Revision sheet: electric potential and capacitors (203-SN2-RE)

This sheet does not redo the course on potential and capacitors, it lists what loses marks on test day. The chapter looks simpler than the one on the field, since everything in it is scalar, and that is precisely the trap: you add without thinking, you confuse a zero value with a zero slope, and you apply to capacitors the rules learned for resistors.

It is written for CEGEP students in the Science program (Sciences de la nature) taking 203-SN2-RE (formerly 203-NYB).

Mark this sheet as read or add it to your favourites: a free account, no password, keeps your read sheets and favourites from one visit to the next, tells you which chapter to tackle next and lets you ask for the method or exercise you are missing. Create your space, an email is enough.

The thread of the chapter

The potential is a SCALAR and it tells you what a displacement costs, never what the force is: $V=0$ does not imply $\vec{E}=\vec{0}$ , and it is the slope of $V$ , not its value, that makes the spark.

This chapter is part of 203-SN2-RE, Electricity and Magnetism

What is on this sheet

1. The essentials
2. The mistakes that cost marks
3. Which method to choose
4. How the answer is expected to be written
5. Check before you hand in
6. The typical problem, taken apart
7. Learn by heart
8. Frequently asked questions

Topics covered

Electric potential
Scalar superposition
Equipotential surfaces
Gradient of the potential
Conductor in electrostatic equilibrium
Point effect
Capacitance of a capacitor
Series and parallel combinations
Stored energy
Energy density of the field
Dielectric strength
Geometry-only capacitance

The essentials

Potential: a scalar that says nothing about the force

• $V=\displaystyle\sum_{i}\dfrac{kq_{i}}{r_{i}}$ , an ALGEBRAIC sum of numbers, signs included. No projection, no components: this is the one place in the chapter where you save work compared with the field.
• $\vec{E}=-\vec{\nabla}V$ : the field depends on how $V$ VARIES around the point, never on its value at that point. With radial symmetry, $E_{r}=-\dfrac{dV}{dr}$ .
• Equipotentials are everywhere perpendicular to the field lines, and crowd together where the field is strong: they read like contour lines.
• Conductor in equilibrium: $\vec{E}=\vec{0}$ in the metal, all the charge on the surface, the whole volume at the SAME potential, and $E=\dfrac{\sigma}{\varepsilon_{0}}$ just outside.

V

vanishes at two points,

E

at a third: at

x=+7

cm, arriving costs nothing but you feel a force; at

x=+13.3

cm, it is the opposite. Field arrows not to scale.

The parallel-plate capacitor: uniform field, parallel equipotentials

• Two plates of area $A$ at distance $d$ , carrying $+Q$ and $-Q$ : between them the field is uniform, $E=\dfrac{\Delta V}{d}=\dfrac{\sigma}{\varepsilon_{0}}$ , and zero outside if you neglect the edges.
• The equipotentials are planes parallel to the plates and evenly spaced: the potential drops linearly from the $+$ plate to the $-$ plate.
• $C=\dfrac{\varepsilon_{0}A}{d}$ : widening the plates or bringing them closer increases the capacitance.

Field lines parallel and equally spaced, from

+

-

; grey equipotentials parallel to the plates and equally spaced: from one grey line to the next,

V

drops by the same amount.

Between the plates, the field is calculated by $\dfrac{\Delta V}{d}$ , never by $\dfrac{kq}{r^{2}}$ : it is the same everywhere between the plates, near one plate as at the middle.

The dielectric: less field, more capacitance

• An insulator slid between the plates becomes POLARIZED: bound charges appear on its faces, of sign opposite to the neighbouring plate.
• These charges weaken the field in the insulator: at fixed charge, $E=\dfrac{E_{0}}{\kappa}$ , so $\Delta V$ drops and the capacitance is multiplied by the dielectric constant, $C=\kappa C_{0}$ .
• Dielectric strength is the maximum field before breakdown, $3.00\times 10^{6}$ V/m for air: it is what limits the voltage that can be applied.

In the insulator, most of the lines end on the bound charges of its face (small signs): fewer pass through, so the field there is weaker,

E=\dfrac{E_{0}}{\kappa}

Capacitor disconnected: $Q$ stays fixed, $\Delta V$ and $E$ drop. Capacitor left connected: $\Delta V$ stays fixed, and it is $Q$ that is multiplied by $\kappa$ .

Capacitors: what depends on the geometry and what does not

• $C=\dfrac{Q}{\Delta V}$ depends ONLY on the geometry and the dielectric. Parallel-plate: $C=\dfrac{\varepsilon_{0}A}{d}$ . Isolated sphere: $C=4\pi\varepsilon_{0}R$ . Coaxial: $\dfrac{C}{L}=\dfrac{2\pi\varepsilon_{0}}{\ln(b/a)}$ .
• Combinations, the REVERSE of resistors: in PARALLEL the capacitances add and the voltage is common; in SERIES it is the reciprocals that add and the CHARGE is common.
• Energy: $W=\dfrac{1}{2}C(\Delta V)^{2}=\dfrac{Q^{2}}{2C}=\dfrac{1}{2}Q\,\Delta V$ . Energy density of the field: $u=\dfrac{1}{2}\varepsilon_{0}E^{2}$ , wherever there is a field, capacitor or not.
• Constants: $\varepsilon_{0}=8.854\times 10^{-12}$ F/m, $k=8.988\times 10^{9}$ N·m $^{2}$ /C $^{2}$ , dielectric strength of air $3.00\times 10^{6}$ V/m.

Series, left: the same charge

Q

on each, and the voltages add,

\Delta V=\Delta V_{1}+\Delta V_{2}

. Parallel, right: the same

\Delta V

across each, and the charges add.

The mistakes that cost marks

These are the errors I correct most often in session. Each one costs marks on a paper, even when the reasoning behind it is right.

1. Two charges of opposite signs on an axis. You are asked where the field vanishes, having already found where the potential vanishes.

3 marks, and often the next question with it

What not to write

The potential vanishes at $x=+7.00$ cm, so the field does too: the field is derived from the potential.

What to write

These are two different conditions. $V=0$ requires $\dfrac{q_{1}}{r_{1}}=\dfrac{|q_{2}|}{r_{2}}$ , at $x=+7.00$ cm; $E=0$ requires $\dfrac{q_{1}}{r_{1}^{2}}=\dfrac{|q_{2}|}{r_{2}^{2}}$ , at $x=+13.3$ cm.

Why: The field is the NEGATIVE OF THE GRADIENT of $V$ : it depends on how $V$ varies around the point, not on its value. A function can equal zero with a nonzero slope, like $f(x)=x$ at $0$ . The counterexample to remember: in a charged conductor, $\vec{E}=\vec{0}$ and $V\ne 0$ ; on the perpendicular bisector plane of a dipole, $V=0$ and $\vec{E}\ne\vec{0}$ .

2. A $354$ pF capacitor is charged at $200$ V instead of $100$ V.

2 marks, and the impossibility of handling the rest of the chapter

What not to write

The charge doubles, so $C=\dfrac{Q}{\Delta V}$ doubles too: the capacitance becomes $708$ pF.

What to write

The charge doubles AND the voltage doubles, so the ratio does not move: $C=354$ pF, as before, as when empty.

The charge is proportional to the voltage: doubling one doubles the other, and the slope, which is the capacitance, does not move. An empty capacitor, at the origin, has the same capacitance.

Why: $C=\dfrac{Q}{\Delta V}$ is a DEFINITION of the ratio, not a law of variation. The proof of $C=\dfrac{\varepsilon_{0}A}{d}$ shows it: the $Q$ cancels along the way, and only the geometry is left. A discharged capacitor, with $Q=0$ and $\Delta V=0$ , has exactly the same capacitance as a full one.

3. Three $2.00$ μF capacitors are connected in series.

The whole network question, 4 marks

What not to write

In series the values add, as for resistors: $C_{\text{eq}}=6.00$ μF.

What to write

In series it is the RECIPROCALS that add: $\dfrac{1}{C_{\text{eq}}}=\dfrac{3}{2.00}$ , so $C_{\text{eq}}=0.667$ μF, three times LESS than each one.

Why: The rule is exactly reversed compared with resistors, and memory mixes the two under pressure. The sure way to recover the right direction is GEOMETRIC: putting in parallel amounts to widening the plates, hence increasing $A$ , hence $C$ ; putting in series amounts to stacking the thicknesses, hence increasing $d$ , hence lowering $C$ .

4. Calculate the potential created at the origin by $+5.00$ nC and $-2.00$ nC placed symmetrically at $3.00$ cm.

2 marks, and the sign carries the error into the work asked next

What not to write

$V=\dfrac{k(5.00)}{0.0300}+\dfrac{k(2.00)}{0.0300}=2098$ V, taking the absolute values of the charges as you do for the field.

What to write

$V=\dfrac{k}{0.0300}(5.00-2.00)\times 10^{-9}=899$ V: the SIGN of each charge goes directly into the sum.

Why: For the FIELD, you handle vectors and often reason on magnitudes with a direction decided on a diagram. For the POTENTIAL, there is no direction to decide: the negative charge makes a negative contribution, period. That is the saving the potential offers, and taking absolute values destroys it.

5. Calculate the force between the plates of a charged parallel-plate capacitor, knowing $Q$ and the field $E$ between the plates.

2 marks, with a plausible result that survives proofreading

What not to write

$F=QE=3.54\times 10^{-8}\times 2.00\times 10^{5}=7.08\times 10^{-3}$ N.

What to write

$F=\dfrac{Q^{2}}{2\varepsilon_{0}A}=3.54\times 10^{-3}$ N, HALF as much, or by energy $F=\dfrac{W}{d}$ , which gives the same result.

The two plates attract each other with the same force

F=\dfrac{Q^{2}}{2\varepsilon_{0}A}

. Each one feels only the field created by the OTHER, half of the field

E

drawn between them.

Why: A plate does not feel ITS OWN field: the one acting on it is the field created by the other plate alone, namely $\dfrac{\sigma}{2\varepsilon_{0}}$ , half the total field. The energy method, $F=\dfrac{dW}{dd}$ , avoids having to decide which field acts on what, and it is the one to prefer.

6. Compare the danger of a Van de Graaff generator at $750$ kV and of a wall outlet at $120$ V.

2 marks on a reasoning question

What not to write

The Van de Graaff is six thousand times more dangerous, since the danger can be read off the number of volts.

What to write

The Van de Graaff stores only $\dfrac{1}{2}CV^{2}=7.8$ J with its $28$ pF, and its discharge is brief. The wall outlet delivers amperes indefinitely: it is the one that kills.

Why: What hurts is a CURRENT, its duration and its path, not a voltage. Voltage determines initiation and breakdown, stored energy determines what can be delivered. A $25$ cm sphere at $750$ kV contains the energy of a $1$ kg mass falling about $80$ cm, and that is why the experiment is done in class.

Which method to choose

Which method, according to what the problem gives

First identify the nature of the object: point charges, a continuous distribution, a conductor, or a network of capacitors.

The large sphere takes four fifths of the charge, and yet it is the small one that feels the stronger field: at a common potential,

E=\dfrac{V}{R}

. Surface fields to scale.

If POINT charges and a question about potential or energy → Scalar superposition, $V=\sum\dfrac{kq_{i}}{r_{i}}$ with the signs. For the assembly energy, $U=\sum_{\text{pairs}}\dfrac{kq_{i}q_{j}}{r_{ij}}$ , once per pair.

Example: $+5.00$ nC and $-2.00$ nC at $3.00$ cm from the origin: $V=899$ V.
If A CONTINUOUS distribution on the axis of a ring, a disk or a rod → Integrate the SCALAR $\displaystyle\int\dfrac{k\,dq}{r}$ , which is much simpler than the field, then differentiate to get $E_{r}=-\dfrac{dV}{dr}$ .

Example: Ring of radius $a$ : $V=\dfrac{kQ}{\sqrt{a^{2}+z^{2}}}$ , then $E_{z}=\dfrac{kQz}{(a^{2}+z^{2})^{3/2}}$ .
If A CONDUCTOR, solid, hollow, or connected to another → The field is zero inside and the whole volume is at the same potential. Two connected conductors come to the same $V$ , not to the same charge: $\dfrac{Q_{1}}{Q_{2}}=\dfrac{R_{1}}{R_{2}}$ for spheres.

Example: Spheres of $8.00$ and $2.00$ cm connected, $12.0$ nC in all: $9.60$ and $2.40$ nC, at $1.08$ kV.
If A NETWORK of capacitors and a supply voltage → Reduce from the inside out down to $C_{\text{eq}}$ , calculate $Q_{\text{total}}=C_{\text{eq}}\Delta V$ , then work back down, carrying the CHARGE through the series and the VOLTAGE through the parallels.

Example: Four capacitors at $24.0$ V: $C_{\text{eq}}=4.00$ μF, $Q=96.0$ μC, $W=1.15$ mJ.
If The problem mentions BREAKDOWN, a spark or a maximum voltage → Go through the FIELD, not the voltage: compare $E$ with the dielectric strength. For a sphere, $E=\dfrac{V}{R}$ ; for a parallel-plate capacitor, $E=\dfrac{\Delta V}{d}$ .

Example: Sphere of $25.0$ cm: $V_{\max}=3.00\times 10^{6}\times 0.250=750$ kV.

Two connected conductors share the potential, never the charge. And it is the SMALLEST radius that carries the strongest field, hence that breaks down first.

How the answer is expected to be written

A marker ticks steps. Here they are in order, with the concluding sentence expected word for word.

Solving a network of capacitors

When to use it: Whenever a circuit has more than two capacitors and you are asked for a charge, a voltage or an energy.

1 Describe the topology in one sentence, explicitly naming what is in series and what is in parallel.
2 Reduce from the inside out down to $C_{\text{eq}}$ , writing each intermediate step.
3 Calculate the total charge $Q_{\text{total}}=C_{\text{eq}}\Delta V$ delivered by the source.
4 Work back down the tree: the CHARGE passes unchanged through a series, the VOLTAGE passes unchanged through a parallel.
5 Check with two independent controls: the sum of the voltages of a series gives back the voltage of the block, and the sum of the charges of a parallel gives back the charge of the block.

Concluding sentence

“Since $C_{4}$ is in series with everything else, it carries the total charge $Q_{\text{total}}=96.0$ μC, so $\Delta V_{4}=\dfrac{Q_{\text{total}}}{C_{4}}=8.00$ V.”

The trap: Carrying the voltage through a series or the charge through a parallel: it is the reverse, and the error spreads to every answer that follows.

Marking: 1 mark for the topology, 1 mark for $C_{\text{eq}}$ , 2 marks for the work back down, 1 mark for the two controls.

Check before you hand in

Five minutes of checking recover more marks than one more problem started in a hurry.

An equivalent capacitance in series is SMALLER than the smallest

Compare $C_{\text{eq}}$ with each of the capacitances put in series. If it is larger than one of them, the law of reciprocals was applied wrongly.

Three times $2.00$ μF in series give $0.667$ μF, never $6.00$ μF.
The two balances of the network

Add the voltages of a series branch: they must give back the voltage of the block. Add the charges of the parallel branches: they must give back the total charge.

$5.33+10.7=16.0$ V, and $64.0+32.0=96.0$ μC: both close.
Energy by two routes

Calculate $\dfrac{1}{2}C_{\text{eq}}(\Delta V)^{2}$ , then add the $\dfrac{1}{2}C_{i}(\Delta V_{i})^{2}$ element by element. The two must agree, and this check tests ALL the voltages at once.

$1.15$ mJ from the overall figure, $85.3+171+512+384=1152$ μJ from the detail.
The unit of capacitance

Check the order of magnitude: laboratory capacitors range from the picofarad to the millifarad. The entire Earth makes only $709$ μF.

A capacitance of $3.5$ F obtained for two plates of $20$ cm $^{2}$ signals a power-of-ten error, the correct value being $354$ pF.

The typical problem, taken apart

A network of four capacitors, taken apart then put back together

$C_{1}=6.00$ μF and $C_{2}=3.00$ μF are in series; this branch is in parallel with $C_{3}=4.00$ μF; the whole is in series with $C_{4}=12.0$ μF. The entire circuit is supplied at $\Delta V=24.0$ V. Calculate the charge and the voltage of each capacitor, then the total energy.

The network of the example:

C_{1}

and

C_{2}

in series on the top branch, this branch in parallel with

C_{3}

, and the whole in series with

C_{4}

24.0

Step 1

$\dfrac{1}{C_{12}}=\dfrac{1}{6.00}+\dfrac{1}{3.00}=\dfrac{1}{2.00}$ , so $C_{12}=2.00$ μF

Why

You always reduce from the INSIDE out, starting with the deepest block. Attacking $C_{4}$ first leads nowhere, because you do not yet know what is facing it.

Step 2

$C_{123}=C_{12}+C_{3}=2.00+4.00=6.00$ μF, then $C_{\text{eq}}=\dfrac{6.00\times 12.0}{18.0}=4.00$ μF

Why

A parallel adds, a series follows the law of reciprocals. Immediate check: $C_{\text{eq}}=4.00$ μF is indeed below $6.00$ and $12.0$ , as any series must be.

Step 3

$Q_{\text{total}}=C_{\text{eq}}\Delta V=4.00\times 24.0=96.0$ μC, so $\Delta V_{4}=\dfrac{96.0}{12.0}=8.00$ V and $\Delta V_{\text{block}}=24.0-8.00=16.0$ V

Why

$C_{4}$ is in series with everything else, so it carries the TOTAL charge: that is the property that unlocks the work back down. Cross-check: $\dfrac{96.0}{6.00}=16.0$ V, the same block voltage by the other route.

Step 4

$Q_{3}=C_{3}\Delta V_{\text{block}}=4.00\times 16.0=64.0$ μC and $Q_{1}=Q_{2}=C_{12}\Delta V_{\text{block}}=2.00\times 16.0=32.0$ μC

Why

In a parallel, it is the VOLTAGE that is common, so each branch takes the charge its capacitance dictates. In the series $C_{1}C_{2}$ , it is the CHARGE that is common: both carry $32.0$ μC.

Step 5

$\Delta V_{1}=\dfrac{32.0}{6.00}=5.33$ V and $\Delta V_{2}=\dfrac{32.0}{3.00}=10.7$ V, and their sum does come to $16.0$ V

Why

The SMALLEST capacitor of the series takes the LARGEST voltage, since $\Delta V=\dfrac{Q}{C}$ at equal charge. That is why it is always the smallest one that breaks down first in a series, a design point that a test question likes to bring out.

Step 6

$W=\dfrac{1}{2}C_{\text{eq}}(\Delta V)^{2}=\dfrac{1}{2}\times 4.00\times 24.0^{2}=1152$ μJ, and the detail gives $85.3+171+512+384=1152$ μJ

Why

The energy balance is the final check: it tests all the voltages at once, and a one percent discrepancy exposes a wrong voltage somewhere in the work back down.

The conclusion, written out

$C_{\text{eq}}=4.00$ μF, $Q_{\text{total}}=96.0$ μC, with $8.00$ V across $C_{4}$ , $16.0$ V across $C_{3}$ , $5.33$ V across $C_{1}$ and $10.7$ V across $C_{2}$ . Total energy $1.15$ mJ, checked by two routes.

The classic mistake on this problem: Carrying the block voltage into the series branch $C_{1}C_{2}$ , that is, giving $16.0$ V to each: this mixes up the parallel rule with the series rule, and the total energy gives the error away immediately.

Learn by heart

• $V=\sum\dfrac{kq_{i}}{r_{i}}$ with the SIGNS, and $\vec{E}=-\vec{\nabla}V$ . The value of $V$ says nothing about $\vec{E}$ , only its slope matters.
• Conductor in equilibrium: $\vec{E}=\vec{0}$ inside, charge on the surface, uniform potential, $E=\dfrac{\sigma}{\varepsilon_{0}}$ just outside.
• Two connected conductors: same POTENTIAL, not the same charge. For spheres, $\dfrac{Q_{1}}{Q_{2}}=\dfrac{R_{1}}{R_{2}}$ and $E=\dfrac{V}{R}$ .
• $C=\dfrac{\varepsilon_{0}A}{d}$ for a parallel-plate capacitor, $4\pi\varepsilon_{0}R$ for an isolated sphere. Capacitance NEVER depends on the charge.
• Parallel: the $C$ add, common voltage. Series: the $\dfrac{1}{C}$ add, common charge. The reverse of resistors.
• $W=\dfrac{1}{2}C(\Delta V)^{2}=\dfrac{Q^{2}}{2C}$ , and $u=\dfrac{1}{2}\varepsilon_{0}E^{2}$ in J/m $^{3}$ . Force between plates: $\dfrac{Q^{2}}{2\varepsilon_{0}A}$ , half of $QE$ .
• Air breaks down at $3.00\times 10^{6}$ V/m: a criterion on the FIELD, never on the voltage.

Frequently asked questions

Is the electric field zero where the potential is zero?

No, these are two different conditions. The field is the negative of the gradient of the potential: it depends on how the potential varies around the point, not on its value. On the perpendicular bisector plane of a dipole the potential is zero and the field is at its maximum; in a charged conductor, it is the reverse, the field is zero and the potential is not.

Does the capacitance of a capacitor depend on the charge you put on it?

No. Capacitance depends only on the geometry and the dielectric: area of the plates, distance between them, nature of the insulator. Charging more increases the charge AND the voltage in the same proportion, so their ratio does not change. A discharged capacitor has exactly the same capacitance as a full one.

How do I combine capacitors in series and in parallel?

It is the reverse of resistors. In parallel, the capacitances add and the voltage is common to all branches. In series, it is the reciprocals of the capacitances that add, and the charge is common. A sure way to remember: putting in parallel widens the plates, putting in series stacks the thicknesses.

Why does a point produce a spark before a ball does?

Because two connected conductors are at the same potential, and the field at the surface of a sphere equals the potential divided by the radius. A small radius of curvature therefore produces, for a modest potential, an enormous field, which ionizes the surrounding air. This is the point effect, and it is the principle of the lightning rod.

Where is the energy of a capacitor stored?

In the electric field itself, including in the vacuum between the plates where there is no charge at all, with a density equal to half the permittivity of free space times the square of the field. Integrating this density over the volume gives back exactly half the capacitance times the square of the voltage, to the last digit.

Is a voltage of 750,000 volts necessarily deadly?

No. What hurts is a current, its duration and its path through the body, not a voltage. A Van de Graaff generator at 750 kilovolts stores only about eight joules and delivers only a very brief discharge. A household outlet at 120 volts, on the other hand, can deliver several amperes indefinitely, and it is the one that is dangerous.

Practise it

Corrected exercises: Electric potential and capacitors, 203-SN2-RE

A method is proved on a paper, not on a sheet. The set for the same chapter takes each of these traps into a problem, with the solution written out step by step.

10 corrected exercises
100 points
150 minutes

Do the exercises

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© Ahmed Squalli Houssaini. Revision sheet published at www.letuteurscientifique.ca/en/fiches/cegep-phys2-potential-capacitors. Free for personal and classroom use; republishing it elsewhere requires written permission (legal notice).

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Revision sheet: electric potential and capacitors (203-SN2-RE)

The essentials

Potential: a scalar that says nothing about the force

The parallel-plate capacitor: uniform field, parallel equipotentials

The dielectric: less field, more capacitance

Capacitors: what depends on the geometry and what does not

The mistakes that cost marks

1. Two charges of opposite signs on an axis. You are asked where the field vanishes, having already found where the potential vanishes.

2. A 354354354 pF capacitor is charged at 200200200 V instead of 100100100 V.

3. Three 2.002.002.00 μF capacitors are connected in series.

4. Calculate the potential created at the origin by +5.00+5.00+5.00 nC and −2.00-2.00−2.00 nC placed symmetrically at 3.003.003.00 cm.

5. Calculate the force between the plates of a charged parallel-plate capacitor, knowing QQQ and the field EEE between the plates.

6. Compare the danger of a Van de Graaff generator at 750750750 kV and of a wall outlet at 120120120 V.

Which method to choose

Which method, according to what the problem gives

How the answer is expected to be written

Solving a network of capacitors

Check before you hand in

The typical problem, taken apart

A network of four capacitors, taken apart then put back together

Learn by heart

Frequently asked questions

Is the electric field zero where the potential is zero?

Does the capacitance of a capacitor depend on the charge you put on it?

How do I combine capacitors in series and in parallel?

Why does a point produce a spark before a ball does?

Where is the energy of a capacitor stored?

Is a voltage of 750,000 volts necessarily deadly?

Corrected exercises: Electric potential and capacitors, 203-SN2-RE

See also

Looking for a 203-SN2-RE tutor in Montreal?

2. A $354$ pF capacitor is charged at $200$ V instead of $100$ V.

3. Three $2.00$ μF capacitors are connected in series.

4. Calculate the potential created at the origin by $+5.00$ nC and $-2.00$ nC placed symmetrically at $3.00$ cm.

5. Calculate the force between the plates of a charged parallel-plate capacitor, knowing $Q$ and the field $E$ between the plates.

6. Compare the danger of a Van de Graaff generator at $750$ kV and of a wall outlet at $120$ V.