Electrostatics and circuits: the test

Electrostatics arrives at the liceo scientifico (the Italian science-focused high school) between January and March of Year 4 (at the classical, human-sciences and languages high schools, a year later, in September-December of Year 5), and opens the block that will fill the rest of the physics course up to the maturità. The test is almost always in two parts: one of pure electrostatics — charges, field, potential — and one of direct-current circuits, with resistors to reduce.

The problem I see most often is not the calculation. It is that in three weeks four quantities arrive whose names are similar and which behave differently — force, field, potential energy, potential — and nobody puts them in a row next to each other. If you do not make that table, you confuse $E$ and $V$ all year, and at the maturità you pay for it again.

Here you will find the four quantities lined up, the method for reducing a circuit, three worked examples with the calculations in full and the five mistakes that cost the most marks.

Coulomb: the same form as gravitation, two differences

F = k \frac{∣ q _{1} ∣ ∣ q _{2} ∣}{r ^{2}}, k = \frac{1}{4 π ε _{0}} \approx 8.99 \times 1 0^{9} N m^{2} / C^{2}

The structure is identical to that of the law of universal gravitation: product of the “charges” divided by the square of the distance. There are two differences, and both are material for a theory question.

The electric force can be attractive or repulsive, the gravitational one only attractive. That is why the formula uses the magnitudes of the charges: you decide the direction from the drawing, looking at the signs, not from the result of the calculation. If you put the signs inside the formula you get negative forces that you cannot interpret.

It is enormously stronger. Between two protons the electric force is about $1 0^{36}$ times the gravitational one. That is why in electrostatics problems the weight of the particles is always neglected, unless the question says explicitly otherwise.

The four quantities, in a row

This is the table worth having in front of you throughout the chapter. The first two depend on the test charge, the other two do not.

Quantity	Symbol	Vector or scalar?	Formula (point charge)	Unit
Force	$F$	vector	$F = k \frac{∣ q _{1} q _{2} ∣}{r ^{2}}$	N
Electric field	$E$	vector	$E = k \frac{∣ Q ∣}{r ^{2}}$	N/C = V/m
Potential energy	$U$	scalar	$U = k \frac{q _{1} q _{2}}{r}$	J
Potential	$V$	scalar	$V = k \frac{Q}{r}$	V

There are only two relations between them, and they are worth reading as definitions:

E = \frac{F}{q}, V = \frac{U}{q}

In words: field and potential describe the space, force and energy describe what happens to a charge placed in it. The field exists even if there is no test charge; the force does not.

The two differences to keep in mind so as not to confuse $E$ and $V$ : the field is a vector and goes as $1/ r^{2}$ , the potential is a scalar and goes as $1/ r$ . And in the potential the sign of the charge goes inside: a negative charge generates a negative potential. In the field it does not, you use the magnitude and the direction comes from the drawing (outward from positive charges, inward towards negative ones).

The work done by the electric force to move a charge from A to B is

L_{A B} = q (V_{A} - V_{B}) = - Δ U

and in a uniform field — the one between the plates of a parallel-plate capacitor — the formulas simplify:

E = \frac{Δ V}{d}, L = q E d

Capacitors

C = \frac{Q}{Δ V}, C_{parallel-plate} = ε_{r} ε_{0} \frac{S}{d}, W = \frac{1}{2} C Δ V^{2}

Capacitance is measured in farads, and the farad is huge: in problems you find microfarads ( $1 0^{- 6}$ F) and picofarads ( $1 0^{- 12}$ F). The dielectric between the plates increases the capacitance by a factor $ε_{r}$ , which is always greater than 1.

The trap lies in the connections, because for capacitors the formulas are swapped compared with resistors:

parallel: C_{eq} = C_{1} + C_{2}, series: \frac{1}{C _{eq}} = \frac{1}{C _{1}} + \frac{1}{C _{2}}

It is not the textbook being nasty: in parallel the plates side by side offer more surface, and more surface means more capacitance.

Current, Ohm and Joule

i = \frac{Δ q}{Δ t}, Δ V = R i, R = ρ \frac{l}{S}, P = Δ V i = R i^{2}

The first Ohm’s law ( $Δ V = R i$ ) links voltage and current; the second ( $R = ρl / S$ ) says what resistance depends on: material, length and cross-section. A longer wire resists more, a thicker one less — the analogy with water in a pipe really works here.

For resistors:

series: R_{eq} = R_{1} + R_{2}, parallel: \frac{1}{R _{eq}} = \frac{1}{R _{1}} + \frac{1}{R _{2}}

The criterion for recognising them in a diagram, without looking at how it is drawn:

in series if they carry the same current, that is, if there is no node between one and the other;
in parallel if they have the same voltage across them, that is, if they are connected to the same two nodes.

Two checks that save the exercise: the equivalent resistance of a series is always greater than the largest in the group; that of a parallel is always smaller than the smallest. If you get the opposite, you have used the wrong formula.

Three worked examples

1. Force between two charges

Two point charges $q_{1} = + 3.0 μ C$ and $q_{2} = - 2.0 μ C$ are at a distance $r = 0.20$ m. Calculate the force between them.

First the conversions: $3.0 μ C = 3.0 \times 1 0^{- 6}$ C and $2.0 μ C = 2.0 \times 1 0^{- 6}$ C.

F = 8.99 \times 1 0^{9} \cdot \frac{3.0 \times 1 0 ^{- 6} \cdot 2.0 \times 1 0 ^{- 6}}{( 0.20 ) ^{2}} = 8.99 \times 1 0^{9} \cdot \frac{6.0 \times 1 0 ^{- 12}}{0.040}

F \approx 1.3 N

The charges have opposite signs, so the force is attractive: each is directed towards the other. This sentence has to be written — it is half the answer, and the calculation alone does not contain it.

2. Field and potential generated by a charge

A charge $Q = + 5.0$ nC. Find the electric field and the potential at a distance $r = 0.10$ m.

E = 8.99 \times 1 0^{9} \cdot \frac{5.0 \times 1 0 ^{- 9}}{( 0.10 ) ^{2}} \approx 4.5 \times 1 0^{3} N/C

V = 8.99 \times 1 0^{9} \cdot \frac{5.0 \times 1 0 ^{- 9}}{0.10} \approx 4.5 \times 1 0^{2} V

The field is directed radially outwards, because the charge is positive; the potential is a number, positive, with no direction.

Note the ratio between the two values: $V = E \cdot r$ for a point charge, and it is a quick check. But careful, it holds only in this case: in a uniform field the relation is $Δ V = E d$ with $d$ the distance between the plates, and in general the link is more complicated.

3. Circuit with series and parallel

A $12$ V generator supplies $R_{1} = 6.0 Ω$ in series with the parallel combination of $R_{2} = 6.0 Ω$ and $R_{3} = 3.0 Ω$ . Find the total current, the voltage across each resistor and the power dissipated.

First the parallel, which is the “innermost block”:

R_{23} = \frac{R _{2} \cdot R _{3}}{R _{2} + R _{3}} = \frac{6.0 \cdot 3.0}{9.0} = 2.0 Ω

The check works: $2.0 Ω$ is smaller than the smallest of the two, that is $3.0 Ω$ .

Now the series:

R_{tot} = R_{1} + R_{23} = 6.0 + 2.0 = 8.0 Ω

i = \frac{Δ V}{R _{tot}} = \frac{12}{8.0} = 1.5 A

This current flows through the whole of $R_{1}$ , so

Δ V_{1} = R_{1} i = 6.0 \cdot 1.5 = 9.0 V

and across the parallel there remains $12 - 9.0 = 3.0$ V, the same across both resistors:

i_{2} = \frac{3.0}{6.0} = 0.50 A, i_{3} = \frac{3.0}{3.0} = 1.0 A

The sum is $1.5$ A, that is the total current: the calculation is consistent. Finally the power:

P = Δ V \cdot i = 12 \cdot 1.5 = 18 W

Note that in the parallel the smaller resistance takes the larger current. It is the quickest plausibility check on this type of exercise.

The five mistakes that cost the most marks

1. Confusing field and potential. $E$ is a vector that goes as $1/ r^{2}$ , $V$ is a scalar that goes as $1/ r$ . If you add two fields as if they were numbers, you are wrong: fields add vectorially, potentials algebraically. This is by far the most expensive mistake in the chapter.

2. Confusing potential and potential energy. $V$ is a property of the point in space, $U$ also depends on the charge you put in it: $U = q V$ . A point has a potential even if there is no charge.

3. Swapping the series and parallel formulas for capacitors. They are the opposite of those for resistors. It is the mistake the test deliberately checks with a short exercise.

4. Forgetting to invert at the end of the parallel. With $1/ R_{eq} = 1/6 + 1/3 = 1/2$ the answer is $R_{eq} = 2 Ω$ , not $0.5 Ω$ . The “smaller than the smallest” check always catches it.

5. Not converting the submultiples. $μ C$ , nC, pF, k $Ω$ , mA: the conversions must be done before touching the calculator, and the distance must be squared after being converted to metres.

Other set-up mistakes that recur throughout high school physics — signs, units, reading the question — are collected in typical physics mistakes at high school.

The two weeks before the test

Redo the table of the four quantities on a blank sheet, every day for a week. Columns: vector or scalar, how it depends on $r$ , unit of measurement. Two minutes, and it is the exercise that pays off most.
Do five circuit reductions without calculating anything: just write which resistors are in series and which in parallel, starting from the innermost block. The wrong reduction is the number-one cause of lost exercises.
Do five Coulomb problems with more than two charges, which are the ones where you have to add vectorially and where you lose most time if you are not trained.
In the last two days a timed test, and mark it by sorting each mistake into three categories: wrong formula, wrong conversion, wrong set-up. They are three different problems with three different remedies.

If you need more practice on field, potential and capacitors, I wrote it up separately in electric potential and capacitors. For circuits with more than one generator the Thevenin and Norton method is in Thevenin and Norton with exercises, and the technical-school version of Ohm and Kirchhoff is in Ohm and Kirchhoff.

How we work on this in lessons

The first thing I do on this chapter is the table of the four quantities, built together with the student, not handed over ready-made. Then we come back to it in every exercise: “is this a vector or a scalar? does it depend on the test charge?”. They are two questions that become automatic in three lessons and on their own close half the mistakes in the test.

On circuits I work with the redrawn diagram: almost all the mistakes come from reading the textbook drawing instead of rewriting the circuit by grouping the blocks. Redrawing it costs a minute and is worth the exercise. Lessons are online and one-to-one: rates and packages are on the pricing page, and the 10-hour package is worth it if you want to cover the whole electrical block rather than a single test.

The quickest way to find out where you stand is to try the mock test on electrostatics and circuits: fifteen questions in fifty minutes, with a breakdown of weak points by topic at the end. The correct answer to every question is free and without registration; the guided step-by-step solution unlocks with the first lesson. If after the test you see that electrostatics holds up but circuits do not (or the other way round), write to me for a free introductory call: we look at the syllabus of your test and I tell you honestly how many lessons you need. The full picture of how I set up the course is on the page about physics tutoring.

Electrostatics and circuits: preparing the Year 4 test