17.3 GAUSS' LAW
17.3.1 Electric Flux
1. Concept of flux
Suppose we place a flat surface of area 1 cm² in an electric field produced by an isolated point charge Q, as shown in Figure 17.9. The surface is ‘immersed’ inside a region full of electric field lines. We orientate the surface until it is perpendicular to the electric field lines. Now we have a certain number of electric field lines ‘flowing’ perpendicularly through the surface. If the surface is closed to the point charge Q, there will be many lines flowing through the surface. Conversely, if the surface is far away from Q, then there will be very few lines flowing through the surface. This concept of some ‘number of lines’ flowing through the surface is known as flux.
|
| Figure 17.22 A flat surface with an area of 1 cm² is placed in an electric field produced by an isolated point charge Q |
2. Electric flux
(a) Surface area in electric field
Consider a small flat surface KLMNK of area Aplaced in an electric field. The surface makes an angle θwith the line JK, which is perpendicular to the electric field lines, as shown in Figure 17.22(a). JKLHJ forms the flat surface whose area $A_n$ is perpendicular to the electric field. Actually $A_n$ is the component of area A, given by
$A_n = A \cos \theta$
(b) Number of field lines flowing through $A_n$
The number of field lines flowing through the normal area $A_n$ depends on the following quantities:
(i) The electric field strength E.
More field lines flow through $A_n$ in a strong electric field strength than in a weak field.
(ii) The size of the surface area $A_n$.
More field lines flow through a large area than through a small area.
(c) The electric flux Φ
The electric flux Φ is a measure of the number of field lines that flow through the normal area $A_n$. Its value has to be determined by the values of E and $A_n$. It is given by
$\Phi = EA_n$
or
$\Phi = EA \cos \theta$
3. Electric flux in terms of vectors
Suppose the field lines flow through a small elemental surface area at angle $\theta$ to the normal, as shown in Figure 17.22(b). We have the following vectors at the surface:
(a) The electric field strength $\theta$ is a vector whose direction points at an angle $\theta$ to the normal of the small surface area.
(b) We can attach a unit vector n to the elemental surface area that points in a direction that is parallel to the normal. Then the small area of magnitude $\Delta A$ may be considered to be a vector $\Delta A$ whose direction is parallel to the direction of unit vector $n$. We can express this elemental area vector as
$\Delta A = \Delta A_n$
Now the electric flux $\Phi$ can be written as a dot product of the $E$ vector and the $\Delta A$ vector, as follows:
$\Phi = E \cdot \Delta A$
4. Quantity
Electric flux is a scalar quantity.
5. Unit of Φ
Electric flux has a unit of
$N \cdot m^2 \cdot C^{-1}$
17.3.2 Electric Flux Through a Large Surface
1. Consider a large surface that is found in a non-uniform electric field. We wish to determine the electric flux that flows through this large surface.
2. Since the electric field is not uniform, the electric field strength Eon one part of the surface may be larger or smaller than the strength at another part of the surface. In other words, the electric field strength varies from point to point on the surface. As a result, the electric flux flowing through one small surface area may be larger or smaller than that flowing through another small surface of equal area.
|
| Figure 17.23 To determine the electric flux flowing through the entire surface, we divide the large surface into a very large number of tiny surfaces |
3. To determine the electric flux flowing through the entire surface, we divide the large surface into a very large number of tiny surfaces as shown in Figure 17.23. Each surface is so small that:
(a) it may be assumed to be flat
(b) the electric field strength at a point on it is assumed to be constant
4. Suppose we divide the large surface into n tiny surfaces. Consider one tiny surface of area ΔA₁ where the electric field strength is E₁. Let the angle between E₁ and the normal to this surface be θ₁. The electric flux through this tiny surface is
$\Delta \Phi_1 = E_1 \Delta A_1 \cos \theta_1$
The electric field strength at a point on another tiny surface of area $\Delta A_2$ has magnitude $E_2$, pointing at an angle $\theta_2$ to the normal of the second tiny surface. The electric flux $\Delta \Phi_2$ flowing through the second tiny surface is given by
$\Delta \Phi_2 = E_2 \Delta A_2 \cos \theta_2$
Using the same prosedure, we determine the electric flux flowing through the rest of the tiny surfaces and obtain electric flux $\Delta \Phi_3$, $\Delta \Phi_4$, $\Delta \Phi_5$, ..., $\Delta \Phi_n$. The electric flux $\Phi$ flowing throught the entire large surface is given by
$\Phi = \Delta \Phi_1 + \Delta \Phi_2 + \Delta \Phi_3 + ... + \Delta \Phi_n$
$= \sum_{i=1}^{n} \Delta \Phi_i$
$= \sum_{i=1}^{n} (E_i \Delta A_i \cos \theta_i)$
5. We may express the electric flux in terms of vectors, as follows:
$\Phi = \sum_{i=1}^{n} (\mathbf{E}_i \Delta \mathbf{A}_i)$
If we let $\Delta A \to 0$, we will obtain
$\Phi = \int \mathbf{E} \cdot d\mathbf{A}$
17.3.3 Gauss' Law
1. Definition
Gauss' law states that:
The total electric flux passing through any closed surface is equal to the total charge enclosed divided by the permittivity of free space.
Mathematically,
$\Phi = \frac{Q}{\epsilon_0}$
where $\epsilon_0$ is the permittivity of free space.
|
| Figure 17.24 The charge enclosed within the surface may be a single point charge, a collection of point charges, or a distribution of excess charges carried by a large object |
2. Elaboration
(a) The surface is known as a Gaussian surface.
(b) It may be a real surface or just an imaginary one.
(c) The shape of the surface is arbitrary. It may take the shape of a sphere, a cubic box, an oval-shaped box, and so on.
(d) The charge enclosed within the surface may be a single point charge, a collection of point charges, or a distribution of excess charges carried by a large object (Figure 17.24).
(e) The charge Q enclosed by the surface is the total charge.
(f) The volume of space enclosed within the Gaussian surface (where charge Q is found) is considered to be ‘inside’ the surface.
(g) If the total charge is positive, the flux flows outwards (away from the charge) through the surface. Conversely, if the total charge is negative, the flux flows inwards (towards the charge).
(h) The unit vector n that is assigned to an elemental area vector ΔA found on the Gaussian surface is assumed to point ‘outward’, or to the ‘outside’.
EXAMPLE 17.10
Determine the total electric flux flowing through a Gaussian surface that encloses completely the following three charges: $q_1 = +2 \mu C$, $q_2 = -15 \mu C$ and $q_3 = +8 \mu C$. State the direction of flow of the electric flux.
Answer
$\Phi=\frac{Q}{\epsilon_0}$
$\Phi=\frac{q_1+q_2+q_3}{\epsilon_0}$
$=\frac{[(2)+(15)+(-8)] \times 10^{-6}}{8.85 \times 10^{-12}}$
$=-5.6 \times 10^5 \ Nm^2 C^{-1}$
Since the total charge is negative, the electric flux flows inwards, through the Gaussian surface towards the collection of charges.
17.3.4 Application of Gauss’ Law
1. Deriving Coulomb’s law using Gauss’ law
(a) Consider two positive point charges in free space, Q1 and Q2, separated by a distance r.
(b) Enclose charge Q1 within an imaginary spherical Gaussian surface of radius r. Place Q1 at the centre of the sphere. Then Q2 will lie on the surface of the sphere.
(c) Since the electric field is a radial field, the electric field lines originate from the positive charge at the centre of the sphere and flow outwards through the spherical surface.
(d) Because each field line is straight, the line ‘pierces’ through the sphere at an angle of 90° to the surface. This means that the electric field strength vector E at any point of the sphere points outwards and is perpendicular to the sphere (along the radius of the sphere).
(e) Divide the sphere into n very small elemental areas. Consider one such small area ΔA1. The vector associated with this area is given by ΔA1n, where n is the unit vector that points outwards, in the same direction as that of vector E. In other words, n and E are parallel to each other (θ=0°).
(f) The electric flux ΔΦ1 that flows outwards through this area is given by
$\Delta \Phi = E_1 \Delta A_1$
(g) Using the same procedure, determine the electric flux $\Delta \Phi_2$, $\Delta \Phi_3$, $\Delta \Phi_4$, ..., $\Delta \Phi_n$ that pass through areas $\Delta A_2$, $\Delta A_3$, $\Delta A_4$ ..., $\Delta A_n$
(h) Find the total electric flux Φ that passes through the entire spherical surface. We get
$\Phi=\Delta \Phi_1+\Delta \Phi_2+\Delta \Phi_3+...+\Delta \Phi_n$
$\Phi=E_1 \Delta A_1+E_2 \Delta A_2+E_3 \Delta A_3+...+E_n \Delta A_n$
But the electric field strength at any point on the spherical surface has the same magnitude, E, since every point is at the same distance r from the point charge Q1. Hence we have
$\Phi=E(\Delta A_1+\Delta A_2+\Delta A_3+...+\Delta A_n)$
$\Phi=E(\text{total area of spherical surface})$
$\Phi=E(4\pi r^2)$
(i) Using Gauss’s law, we have
$\Phi=\frac{Q_1}{\epsilon_0}$
Hence,
$\Phi=E(4\pi r^2)=\frac{Q_1}{\epsilon_0}$
or, $E=\frac{Q_1}{4\pi \epsilon_0 r^2}$
(j) The electric force F acting on Q2 that lies on the spherical surface is given by
$F=Q_2E$
$F=\frac{Q_1Q_2}{4\pi \epsilon_0 r^2}$
This equation is Coulomb’s law.
2. Determining the electric field strength at a point outside an isolated charged spherical conductor
(a) Consider an isolated spherical conductor of radius R that carries an excess positive charge Q uniformly distributed on the surface. The electric field lines originate from the surface and flow outwards perpendicular to the surface, as shown in Figure 17.25.
(b) Enclose the conductor within an imaginary spherical Gaussian surface of radius (r > R). Place the centre of the conductor at the centre of the Gaussian sphere.
|
| Figure 17.25 The electric field lines originate from the surface and flow outwards perpendicular to the surface |
(c) Divide the imaginary sphere into a very large number of small elemental areas. Using the same procedure as described in section (a) above, we get
$E=\frac{Q}{4\pi \epsilon_0 r^2}$ (outside conductor)
Note: E = 0 inside the conductor.
3. Determining the electric field strength at a point outside an isolated long charged cylindrical conductor
(a) Consider an isolated long straight positively charged cylindrical conductor of radius R. It carries a linear charge density of $\lambda C m^{-1}$. The electric field lines originate from the surface of the conductor and flow outwards perpendicular to the surface, as shown in Figure 17.26.
|
| Figure 17.26 The electric field lines originate from the surface of the conductor and flow outwards perpendicular to the surface |
(b) Enclose a portion of the conductor within an imaginary cylindrical Gaussian surface of length L and radius (r > R). Place the axis of the cylindrical conductor on the axis of the Gaussian cylinder.
(c) Divide the curved part of the imaginary cylindrical surface into n very small elemental areas. Consider one such small area ΔA1. The vector associated with this area is given by ΔA1 n, where n is the unit vector that points outwards. Since the field lines are perpendicular to the Gaussian surface, n and the electric field strength vector E are parallel to each other at any point on the surface (θ = 0°).
(d) The electric flux ΔΦ1 that flows outwards through this area is given by
$\Delta \Phi_1 = E_1 \Delta A_1$
$\Delta \Phi_1 = E_1 \Delta A_1 \ cos \ \theta$
(e) Using the same procedure, determine the electric flux $\Delta \Phi_2$, $\Delta \Phi_3$, $\Delta \Phi_4$, ..., $\Delta \Phi_n$ that pass through areas $\Delta A_2$, $\Delta A_3$, $\Delta A_4$ ..., $\Delta A_n$
(f) Find the total electric flux Φ that passes through the entire imaginary cylindrical surface. We get
Φ = (total flux through curved surface) + (total flux through the two flat vertical surfaces at both ends)
But the total flux through both the flat ends is zero because the field lines are parallel to these two surfaces. Hence,
$\Phi$ = total flux through curved surface
$\Phi = \Delta \Phi_1 + \Delta \Phi_2 + ... + \Delta \Phi_3$
$\Phi = E_1 \Delta A_1 + E_2 \Delta A_2 + E_3 \Delta A_3 + ... + E_n \Delta A_n$
The electric field strength at any point on the cylindrical surface has the same magnitude,
E, since it is at the same distance r from the axis of the conductor. Hence we have
$\Phi = E(\Delta A_1 + \Delta A_2 + \Delta A_3 + ... + \Delta A_n)$
$\Phi = E(\text{total area of cylindrical surface})$
$\Phi = E(2 \pi rL)$
(g) Using Gauss’s law, we have
$\Phi = \frac{Q}{\epsilon_0}$
Hence,
$\Phi = E(2 \pi rL) = \frac{Q}{\epsilon_0}$
where Q is the total charge on length L of the conductor that is enclosed by the Gaussian surface.
$Q = \lambda L$
Hence,
$E = \frac{\lambda L}{2 \pi \epsilon_0 rL}$
$E = \frac{\lambda}{2 \pi \epsilon_0 r}$ (outside cylinder)
4. Deducting that excess charge resides on the surface of a charged conductor in electrostatic equilibrium
Consider a positively charged solid conductor without cavity that is in electrostatic equilibrium. Suppose we assume that the excess charge is found inside the conductor. Now imagine there is a Gaussian surface that has the same shape as the conductor but is slightly smaller in size so that this surface lies immediately below the surface of the conductor, as shown in Figure 17.27.
|
| Figure 17.27 imagine there is a Gaussian surface that has the same shape as the conductor but is slightly smaller in size so that this surface lies immediately below the surface of the conductor |
The Gaussian surface would enclose entirely the excess charge that we assume it to be in the interior. According to Gauss’s law,
electric flux Φ ∝ charge enclosed by Gaussian surface.
Electric Field in a Hollow Conductor (Gauss’s Law)
Since the surface encloses some charge, there must be certain quantity of electric flux originating from this excess charge flowing out of the Gaussian surface. This implies that an electric field should exist in the interior of the conductor. If this was the case, then the excess charge would be acted upon by electric forces because we have F = qE. The excess charge would be forced to move if they were to stay inside the conductor. But such happening would contradict the fact that the charged conductor is in electrostatic equilibrium.
Hence, we have to conclude that the electric field could not have existed, meaning E = 0 inside the conductor. If this is the case, we must have the electric flux Φ = 0 inside the conductor. Then the volume enclosed by the Gaussian surface cannot contain any excess charge.
For the electric flux to be zero, the excess charge has to lie outside the Gaussian surface. In other words, the excess charge must reside on the surface of the conductor.
5. Deducting that there is no electric field in the cavity of a hollow charged conductor
Figure 17.28
Consider a charged conductor which has a cavity and in electrostatic equilibrium. There is no charge placed inside the cavity. Imagine that there is a Gaussian surface which has the same shape as the cavity but slightly smaller in size so that the surface lies immediately below the cavity walls.
Gauss’s law states that
electric flux Φ ∝ charge enclosed by Gaussian surface.
Since there is no charge placed inside the cavity, the Gaussian surface does not enclose any charge. Hence, electric flux cannot be found inside the cavity. As a result, there is no electric field inside the cavity if no charge is placed inside the cavity.
EXAMPLE 17.11
A uniformly charged solid spherical insulator has a radius of 23.0 cm. The total charge in the volume is 4.2 pC. Find the E at a position of 0.14 m from the center of the sphere.
Answer
EA = qin / ε0
E(4πr²) = qin / ε0
qin = q(4πr³) / (4πR³) = qr³ / R³
E(4πr²) = qr³ / (ε0R³)
E = [q / (4π ε0R³)] r
E = [4.2 × 10-12 / (4π ε0 (0.23)³)] (0.14) = 0.435 N/C
EXAMPLE 17.12
A soccer goal, found is a city park, is made of tubing that supports an odd-shaped hanging net behind the goal, but has a rectangular opening in front. The height of the opening is 2.5 m and the width is 5.2 m.
If a uniform E-field, with a magnitude of 1.2 N/C, passes through the goal from the front to the back, entering at 90° to the plane of the goal opening, what is the flux through the net?
Also, find the flux through the net if the E-field enters the goal at a 60° angle to the plane of the front of the goal.
In both cases, assume that there is no charge found inside the goal itself.
Answer
No charge inside implies no total flux.
Φtotal = 0 = Φnet + Φfront
0 = Φnet + EA cos 180°
Φnet = −1.2(2.5)(5.2) cos 180° = 15.6 Nm²/C
For part 2, the angle between the E-Field and the Area vector would be 30°.
Φnet = −EA cos 150°
Φnet = −1.2(2.5)(5.2) cos 150° = 13.5 Nm²/C
EXAMPLE 17.13
A pyramid with horizontal square base, 6.00 m on each side, and a height of 4.00 m is placed in a vertical electric field of 52.0 N/C. Calculate the total electric flux through the pyramids four slanted surfaces.
Answer
Since the pyramid’s base is horizontal and the electric field is vertical, then the electric flux that goes through the slanted surfaces of the pyramid must also go also through the base (the lines get in through the slanted surface and get out through the base or in from the base and out through the slanted surfaces).
Consider former case, the electric field and the base’s normal are antiparallel, so the flux through the base is then given by:
Φ = EA cos θ
Φ = EA cos 180°
Φ = −52.0 × 6 × 6 = −1.87 × 103 N·m²/C
So the flux through the slanted surfaces is +1.87 × 103 N·m²/C
EXAMPLE 17.14
Shown below you find three charges:
Q1 = −8.94 μC, Q2 = +4.47 μC, and Q3 = −2.32 μC.
(a) Determine the net electric flux through surface A1.
(b) Determine the net electric flux through surface A2.
Figure 17.29 Charges Q₁, Q₂, Q₃ and Q₄ are enclosed within the Gaussian surfaces A₁ and A₂
Answer
(a) Determine the net electric flux through surface A1.
Considering area A1 we have the situation:
Figure 17.30 Charges Q1, Q2, and Q3 are enclosed within the Gaussian surfaces A1
ΦE→A₁ = Qenclosed-A₁ / ε0 = (Q₁ + Q₂ + Q₃) / ε₀
ΦE→A₁ = (−8.94 μC + 4.47 μC − 2.32 μC) / ε₀
ΦE→A₁ = −6.79 μC / (8.85 × 10−12 C²·N−1·m−2)
ΦE→A₁ = −7.67 × 105 N·m²/C
(b) Determine the net electric flux through surface A2.
Considering area A2 we have the situation:
Figure 17.31 Charge Q2 are enclosed within the Gaussian surfaces A2
ΦE→A₂ = Qenclosed-A₂ / ε₀ = Q₂ / ε₀
ΦE→A₂ = 4.47 μC / (8.85 × 10−12 C²·N−1·m−2)
ΦE→A₂ = 5.05 × 105 N·m²/C
17.3.5 Conclusion – Gauss’ Law
Gauss’ Law is one of the fundamental principles of electrostatics that relates electric flux through a closed surface to the total charge enclosed within that surface. It shows that the electric field produced by charges depends on the amount of charge inside the Gaussian surface, not on charges outside it.
This law provides a powerful method for calculating electric fields, especially when the charge distribution has symmetry such as spherical, cylindrical, or planar symmetry. Using Gauss’ Law, important results like the electric field around point charges, charged spheres, cylinders, and planes can be derived.
Gauss’ Law also explains key properties of conductors in electrostatic equilibrium. It shows that the electric field inside a conductor is zero, excess charges reside on the surface of the conductor, and there is no electric field inside a cavity of a hollow conductor if no charge is present.
In summary, Gauss’ Law helps us understand how electric fields behave and provides a systematic way to analyze many electrostatic systems.
17.3.6 Frequently Asked Questions (FAQ)
Gauss’ Law states that the total electric flux through a closed surface equals the enclosed charge divided by the permittivity of free space. Below are common questions students often ask about this topic.
1. What is Gauss’ Law?
Gauss’ Law is a fundamental law of electrostatics that relates the electric flux through a closed surface to the total charge enclosed within that surface.
It helps determine the electric field when the charge distribution has symmetry.
2. What is a Gaussian surface?
A Gaussian surface is an imaginary closed surface used to apply Gauss’ Law.
- Common Gaussian surfaces include:
- Sphere (for point charges)
- Cylinder (for line charges)
- Pillbox (for infinite plane charges)
3. When is Gauss’ Law most useful?
Gauss’ Law is most useful when the charge distribution has high symmetry, such as:
- Spherical symmetry
- Cylindrical symmetry
- Planar symmetry
Without symmetry, calculating the electric field becomes difficult.
4. What is the electric field inside a conductor?
Inside a conductor in electrostatic equilibrium:
E = 0
This happens because free charges move until the internal electric field cancels out.
5. Where does excess charge reside in a conductor?
Excess charge in a conductor always resides on the surface, not inside the conductor.
This result can be derived from Gauss’ Law.
6. Is there an electric field inside a hollow conductor?
If there is no charge inside the cavity, then:
E = 0
This is the principle behind electrostatic shielding and Faraday cages.
7. What are the main applications of Gauss’ Law?
Gauss’ Law is used to determine the electric field of:
- Point charges
- Spherical charge distributions
- Long charged cylinders
- Infinite charged planes
- Conductors and cavities
Post a Comment for "Gauss’ Law: Electric Flux, Formula, Explanation, and Solved Problems"