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Electric charges, field and Gauss's law

Electric charges attract or repel each other at a distance. Three linked ideas describe it: the force between charges (Coulomb's law), the electric field a charge creates around it and the potential, which measures energy per unit charge. For highly symmetric bodies, Gauss's law gives the field with hardly any calculation. The Electric charges and Gauss's law tools solve both cases step by step.

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Coulomb's law

Two point charges exert equal and opposite forces on each other, along the line joining them: repulsive if they have the same sign and attractive if not. Its size is

F = k \frac{∣ q _{1} q _{2} ∣}{r ^{2}} k = \frac{1}{4 π ε _{0}} = 8.988 \times 1 0^{9} N \cdot m^{2} / C^{2}

With several charges the superposition principle holds: the force on a charge is the vector sum of those exerted by each of the others separately. In practice: write each force with its unit vector $\hat{r}$ (from the charge that pushes to the one that receives) and add components.

Forces on

q_{1}

in example 1: their sum is the net force.

Distances go in meters: 3 cm is 0.03 m and squared,

9 \times 1 0^{- 4}

m². Forgetting it gives forces 10 000 times larger.

Electric field

The electric field at a point is the force per unit charge a test charge $q_{0}$ placed there would feel: $E = F / q_{0}$ . It exists even if there is no test charge. That of a point charge points away from it if positive and toward it if negative:

E = k \frac{q}{r ^{2}} \hat{r} E_{P} = i \sum k \frac{q _{i}}{r _{i}^{2}} \hat{r}_{i}

Field lines draw it: they leave positive charges, arrive at negative ones, never cross, and where they are closer together the field is stronger.

Potential and energy

The potential $V$ is potential energy per unit charge. Unlike the field, it is a scalar: the potentials of several charges add with their signs, with no components.

V = k \frac{q}{r} U = k \frac{q _{1} q _{2}}{r _{12}} E = - \nabla V

The energy of a system of charges is the sum of $U$ over all pairs. If it is negative, the set is bound: work is needed to pull it apart. The field points where the potential drops fastest: at a point there can be a field with zero potential (the perpendicular bisector of a dipole, example 2).

Continuous distributions

A charged object (a rod, a ring, a disk) is divided into pieces $d q$ , each a point charge, and their fields are added (integrated): $d q = λ d l$ , $σ d A$ or $ρ d V$ depending on whether it is a line, a surface or a volume. Before integrating, use symmetry: components that cancel in pairs do not need to be computed.

For example, for a ring of radius $R$ and charge $Q$ , at a point on its axis at a distance $x$ from the center:

E = \frac{k Q x}{( x ^{2} + R ^{2} ) ^{3/2}} V = \frac{k Q}{x ^{2} + R ^{2}}

Gauss's law

The flux of the field through a closed surface depends only on the charge it encloses:

\oint E \cdot d A = \frac{Q _{enc}}{ε _{0}}

It is always true, but it is only useful for computing $E$ when the symmetry lets you choose a Gaussian surface on which $E$ is constant and perpendicular (or parallel) to the surface. Then the integral is just $E$ times an area:

The three symmetries of Gauss's law and their Gaussian surface.

Distribution	Field
Point charge or sphere, outside ( $r > R$ )	$E = \frac{k Q}{r ^{2}}$
Solid insulating sphere, inside ( $r < R$ )	$E = \frac{k Q r}{R ^{3}}$
Conducting sphere, inside	$E = 0$
Infinite line	$E = \frac{λ}{2 π ε _{0} r}$
Infinite plane	$E = \frac{σ}{2 ε _{0}}$
Between two opposite plates (capacitor)	$E = \frac{σ}{ε _{0}}$ , cero afuera,zero outside

In a conductor at equilibrium the field inside is zero and all the charge goes to the surface. That is why a metal shell shields its interior (the Faraday cage).

Common mistakes

Adding the sizes of forces or fields instead of their components.
Putting the charge signs in $F = k ∣ q_{1} q_{2} ∣/ r^{2}$ and also in the direction: the sense is decided only once (attraction or repulsion).
Treating potential as a vector, or adding potentials without their sign.
Using Gauss's law without symmetry (a cube, a finite rod): the equality holds, but $E$ cannot be pulled out of the integral.
Taking all the charge instead of the enclosed charge when the point is inside the distribution.

Worked examples

Example 1Net force on a charge

q_{1} = + 5

µC is at the origin,

q_{2} = - 3

µC at (4, 0) cm and

q_{3} = + 2

µC at (0, 3) cm. Find the net force on

q_{1}

Solution

$q_{2}$ attracts $q_{1}$ (opposite signs), toward $+ x$ : $F_{2} = 8.988 \times 1 0^{9} \cdot \frac{5 \times 1 0 ^{- 6} \cdot 3 \times 1 0 ^{- 6}}{0.0 4 ^{2}} = 84.26$ N.
$q_{3}$ repels $q_{1}$ (same sign), toward $- y$ : $F_{3} = 8.988 \times 1 0^{9} \cdot \frac{5 \times 1 0 ^{- 6} \cdot 2 \times 1 0 ^{- 6}}{0.0 3 ^{2}} = 99.86$ N.
Sum: $F = (84.26, - 99.86)$ N, $∣ F ∣ = 84.2 6^{2} + 99.8 6^{2} = 130.7$ N, at $θ = atan2 (- 99.86, 84.26) = - 49.84°$ from $+ x$ .
Third-law check: the net forces on the three charges add up to zero.

F_{1} = (84.26, - 99.86)

N: 130.7 N at −49.84°.

Solve it in Simuveliq

In Engineering → Electric charges, in Examples, «Load configuration» → Three charges in a triangle (net force).
With Charge to analyze = q1, the «Coulomb's law step by step» computes each force with its unit vector and the vector sum; the Net force on each charge table gives 130.7 N at −49.84° and the forces on q2 and q3, which add up to zero with that on q1.

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Example 2Field and potential of a dipole

A dipole has +2 µC at (−3, 0) cm and −2 µC at (3, 0) cm. Find the field and the potential at P = (0, 3) cm.

Solution

Both charges are $r = 3^{2} + 3^{2} = 4.243$ cm from P, so their fields have the same size: $E_{i} = 8.988 \times 1 0^{9} \cdot \frac{2 \times 1 0 ^{- 6}}{0.0424 3 ^{2}} = 9.986 \times 1 0^{6}$ N/C.
That of the positive one points away from it: direction $(1, 1) / 2$ . That of the negative one points toward it: $(1, - 1) / 2$ . The $y$ components cancel and the $x$ ones add: $E = 2 \cdot 9.986 \times 1 0^{6} \cdot 0.7071 = 1.412 \times 1 0^{7}$ N/C, toward $+ x$ .
Potential: $V = k \frac{( + 2 μ C )}{r} + k \frac{( - 2 μ C )}{r} = 0$ . All along the perpendicular bisector of the dipole the potential is zero, even though the field is not.

E_{P} = (1.412 \times 1 0^{7}, 0)

N/C and

V_{P} = 0

Solve it in Simuveliq

«Load configuration» → Dipole (+ and −): it already has point P at (0, 3) cm.
The Field at P card gives 14.12 MN/C and V = 0 V; the «Electric field and potential at point P» block tabulates each charge's contribution, (7.061·10⁶, ±7.061·10⁶), and their sum. Drag P along the bisector and see that V stays at 0.

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Example 3Charged ring: direct integration

A 10 cm radius ring carries 5 nC spread uniformly. Find the field and the potential on its axis, 7 cm from the center.

Solution

All the pieces $d q$ are the same distance from the point, $r = 0.0 7^{2} + 0. 1^{2} = 0.1221$ m.
By symmetry, the components perpendicular to the axis cancel in pairs; what is left is $d E_{x} = k d q x / r^{3}$ , and since $x$ and $r$ do not change, the integral is just $\int d q = Q$ .
$E = \frac{8.988 \times 1 0 ^{9} \cdot 5 \times 1 0 ^{- 9} \cdot 0.07}{0.122 1 ^{3}} = 1.73$ kN/C and $V = \frac{k Q}{r} = 368.1$ V.
A curiosity: the field is zero at the center and largest at $x = R / 2 = 7.07$ cm, right next to the requested point.

E = 1.73

kN/C along the axis and

V = 368.1

Solve it in Simuveliq

In Engineering → Gauss's law, Charge distribution → «Ring · point on its axis», with Q = 5n, R = 10 and x = 7 cm (the button sets them).
The cards give Electric field at P 1.73 kN/C and Potential at P 368.1 V; the «Step by step» explains the symmetry and compares with a numerical integration. Move the x slider to travel along the E chart.

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Example 4Solid insulating sphere: Gauss's law

An insulating sphere of radius 10 cm has 5 nC spread uniformly through its volume. Find the field 6 cm from the center (inside) and at 20 cm (outside).

Solution

Gaussian surface: a concentric sphere of radius $r$ . By symmetry $E$ is radial and constant over it: $\oint E \cdot d A = E \cdot 4 π r^{2}$ .
Inside, only the enclosed charge counts: $Q_{enc} = Q (r / R)^{3} = 5 \cdot 0. 6^{3} = 1.08$ nC, so $E = \frac{k Q _{enc}}{r ^{2}} = \frac{k Q r}{R ^{3}} = 2.696$ kN/C.
Outside, it encloses all the charge and behaves like a point charge: $E = \frac{k Q}{r ^{2}} = \frac{44.94}{0.04} = 1.123$ kN/C.
The field grows linearly up to the surface ( $E_{max} = k Q / R^{2} = 4.494$ kN/C) and then falls as $1/ r^{2}$ .

E (6 cm) = 2.696

kN/C and

E (20 cm) = 1.123

kN/C.

Solve it in Simuveliq

Charge distribution → «Solid insulating sphere (uniform charge)», Q = 5n, R = 10, r = 6 cm (the button sets them). The Electric field at P card gives 2.696 kN/C, and the «Step by step» shows the Gaussian surface, the enclosed charge (1.08 nC) and the result.
Change r to 20 cm: the field becomes 1.123 kN/C and the step by step uses all the charge. The E-versus-r chart peaks at R.

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Electric charges, field and Gauss's law

Coulomb's law

Electric field

Potential and energy

Continuous distributions

Gauss's law

Common mistakes

Worked examples

Example 1Net force on a charge

Solution

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Example 2Field and potential of a dipole

Solution

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Example 3Charged ring: direct integration

Solution

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Example 4Solid insulating sphere: Gauss's law

Solution

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