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Electric FluxGauss’s LawApplicationsSpherical ChargeConductorsCoulomb’s Law14 Original FiguresLaTeX Equations
Coulomb’s law can always be used to calculate the electric field intensity \(\vec{E}\) for any discrete or continuous charge distribution of charges at rest. The sums or integrals might be complicated (and a computer might be needed to evaluate them numerically), but resulting electric field intensity \(\vec{E}\) can always be found.
In this chapter, we discuss an alternative to Coulomb’s law, called Gauss’s law, that provides a more useful and instructive approach to calculating the electric field in the situations having certain symmetries.
The number of situations that can directly be analyzed using Gauss’s law is small, but those cases can be done with extraordinary ease. Although Gauss’s law and Coulomb’s law gives identical results in the cases in which both can be used. Gauss’s law is considered a more fundamental equation than Coulomb’s law. It is fair to say that while Coulomb’s law provides workhorse of electrostatics, Gauss’s law provides the insight.
CHAPTER 29
29.1 Electric Flux
FIGURE 01
Electric flux through a plane surface in a uniform electric field.The number of electric lines of force passing normally through a certain area is called the electric flux. It is measured by the product of area and the component of electric field intensity normal to the area. It is denoted by the symbol \({\Phi}_{e}\).
Consider a surface ‘\(S\)’ placed in a uniform electric field of intensity ‘\(\vec{E}\)’. Let ‘\(A\)’ be the area of the surface. The component of \(\vec{E}\)normal to the area \(A\) is\( E\cos {\theta}\) as shown in the figure below.
The electric flux through the surface \(S\) is given by
\[{\Phi}_{e}=A (E\cos {\theta})\]
\[{\Phi}_{e}=EA\cos {\theta}\]
\[{\Phi}_{e}=\vec{E}.\vec{A}\]
Thus the electric flux is the scalar product of electric field intensity and the vector area. The SI unit of the electric flux is \(\frac{{Nm}^{2}}{C}\).
CHAPTER 29
29.2 Electric Flux through an Irregular Shaped Object
Consider an object of irregular shape placed in a non-uniform electric field. We want to find out the expression of electric flux through this irregular shaped object.
FIGURE 02
Electric flux through an irregular surface divided into small area elements. We divide the surface into n number of small patches having area \({\Delta A}_{1},{\Delta A}_{2},{\Delta A}_{3},\ldots\ldots,{\Delta A}_{n}\). Let\({\vec{E}}_{1},{\vec{E}}_{2},{\vec{E}}_{3},\ldots\ldots,{\vec{E}}_{n}\) are the electric field intensities which makes angle \({\theta}_{1},{\theta}_{2},{\theta}_{3},\ldots\ldots,{\theta}_{n}\) with thenormal to the area elements \({\Delta A}_{1},{\Delta A}_{2},{\Delta A}_{3},\ldots\ldots,{\Delta A}_{n}\), respectively. If \({\Phi}_{1},{\Phi}_{2},{\Phi}_{3}, \ldots\ldots, {\Phi}_{n}\) be the electric flux through \({\Delta A}_{1},{\Delta A}_{2},{\Delta A}_{3},\ldots\ldots,{\Delta A}_{n}\), then the total electric flux \({\Phi}_{e}\)will be:
Where \({\Delta \vec{A}}_{1},{\Delta \vec{A}}_{2},{\Delta \vec{A}}_{3},\ldots\ldots,{\Delta \vec{A}}_{n}\) are the vector areas corresponding to area elements \({\Delta A}_{1},{\Delta A}_{2},{\Delta A}_{3},\ldots\ldots,{\Delta A}_{n}\)respectively.
When\( n\to \infty \), or \(\Delta A\to 0,\) then the sigma is replaced by the surface integral i.e.,
\[{\Phi}_{e}=\int_{S}\vec{E}.d\vec{A}\]
By convention, the outward flux is taken as positive and inward flux is taken as negative.
CHAPTER 29
29.3 Gauss’s Law
Statement
The total electric flux through any close surface is \(\frac{1}{{\epsilon}_{0}}\) times the total charge enclosed by the surface.
Explanation
The Gauss’s law gives the relation between total flux and total charge enclosed by the surface. Consider a collection of positive and negative charges in a certain region of space. According to Gauss’s law:
\[{\Phi}_{e}=\frac{q}{{\epsilon}_{0}} ---- (1)\]
where q is the net charge enclosed by the surface. Also,
If the charge is distributed into a volume having uniform volume charge density ‘\(\rho \)’, then charge enclosed \(q\) by Gaussian surface is described by expression:
\[\oint\vec{E}.d\vec{A}=\frac{q}{{\epsilon}_{0}} where ‘q’ is the total charge enclosed\]
If the charge is uniformly distributed into a volume having charge density ‘ρ’, then
\[{\Phi}_{e}= \oint\vec{E}.d\vec{A}= \frac{1}{{\epsilon}_{0}}\int_{v}\rho dv ---- (1)\]
If the charge is uniformly distributed over a surface having a surface charge density ‘σ’, then
\[{\Phi}_{e}= \oint\vec{E}.d\vec{A}= \frac{1}{{\epsilon}_{0}}\int_{s}\sigma dA ---- (2)\]
Equation (1) and (2) are the integral form of Gauss’s law.
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CHAPTER 29
29.5 Applications of Gauss’s law
Gauss’s law can be used to calculate the electric field intensity due to certain charge distributions if the charge distribution has the greater symmetry.
CHAPTER 29
29.5.1 Electric Field due to Infinite Line of Charge
FIGURE 03
Gaussian cylindrical surface around an infinite line of charge. Consider a section of infinite line of charge having uniform linear charge density ‘\(\lambda \)’ as shown in the figure below.
We want to find out electric field intensity at any point \(‘P’\) which is at distance ‘\(r\)’ from the wire. For this we consider cylindrical Gaussian surface which passes through point \(‘P’\). The electric flux passing through the cylinder is given as
\[{\Phi}_{e}=\oint_{S}\vec{E}.d\vec{A}\]
The surface ‘\(S\)’ of the cylinder consist of three parts i.e., \(S_{1}, S_{2}andS_{3}\), where
\(S_{1}\) = Area of top cross section of cylindrical Gaussian surface
\(S_{2}\) = Area of bottom cross section of cylindrical Gaussian surface
\(S_{3}\) = Area of curved part of Gaussian surface
This expression gives the electric field intensity due to infinite line of charge.
Problem 20. An infinite line of charge produces a field of \(4.52\times {10}^{4}\frac{N}{C}\) at a distance of 1.96 m. calculate the linear charge density?
Sample problem 5. A plastic rod whose length is 220 cm and whose radius is 3.6 mm carries a negative charge q of magnitude \(3.8\times {10}^{-7}C\) spread uniformly over its surface. What is the electric field near the midpoint of the rod at a point on its surface?
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Solution:\(L=220cm=2.2 m\)
\[r=3.6\times {10}^{-3}m\]
\[q=-3.8\times {10}^{-7}C\]
\[E=?\]
As Electric field intensity due to infinite line of charge is:
29.5.2 Electric Field at a Point Due to Infinite Sheet of Charge
FIGURE 04
Gaussian cylinder through an infinite sheet of charge.Consider an infinite sheet of charge having constant surface charge density ‘σ’. The figure shows a small portion of such sheet.
We want to find electric field intensity at point ‘P’ which is at the distance ‘r’ from sheet. For this we consider a cylindrical Gaussian surface as shown in the figure below.
The net electric flux passing through the cylinder is given as
\[{\Phi}_{e}=\int_{S}\vec{E}.d\vec{A}\]
We divide the cylindrical Gaussian surface into three parts i.e., \(S_{1}, S_{2}andS_{3}\), where
\(S_{1}\) = Left cross sectional area of cylindrical Gaussian surface
FIGURE 05
Three parts of the cylindrical Gaussian surface for an infinite sheet.\(S_{2}\) = Right cross sectional area of cylindrical Gaussian surface
\(S_{3}\) = Area of curved of cylindrical Gaussian surface
In case of surfaces \(S_{1}andS_{2}\),\(\vec{E}∥d\vec{A}\) are parallel to each other i.e., θ = 0˚ and \(\left|{dA}_{1}\right|=\left|dA_{2}\right|=dA\).
\[{\Phi}_{e}=\int_{S_{1}}E dA+ \int_{S_{2}}E dA\]
\[{\Phi}_{e}=E \int_{S_{1}}dA+ E\int_{S_{2}} dA∴E is constant\]
This is the expression of electric field intensity due to infinite sheet of charge.
CHAPTER 29
29.5.3 Electric Field due to Spherical Shell of Charge
Question: Show that the uniform spherical shell of charge behaves, for all external points, as if all its charge were concentrated at its center.
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Proof: Consider a thin spherical shell of radius ‘\(R\)’ which have the charge ‘\(q\)’ with constant surface charge density ‘\(\sigma \)’. The surface charge density
\[\sigma = \frac{q}{A}\Rightarrow \sigma = \frac{q}{4\pi r^{2}} ∴A= 4\pi r^{2} (Surface area of sphere) \]
\[\Rightarrow q= 4\pi r^{2} \sigma \]
Consider a point ‘\(P\)’ outside the shell. We want to find out electric field intensity due to this charge distribution. For this we consider a spherical Gaussian surface of radius \(r > R\) which passes through point ‘\(P\)’ as shown in the figure below.
According to Gauss’s law,
FIGURE 06
Spherical Gaussian surface outside a uniformly charged spherical shell.
Thus the uniform spherical shell of charge behaves like a point charge for all the points outside the shell.
Question: Show that the uniform spherical shell of charge exerts no electrostatic force on a charged particle placed inside the shell.
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Consider a point ‘\(P\)’ inside the shell. We want to fine out electric field intensity ‘\(E\)’ at point ‘\(P\)’ due to this symmetrical charge distribution. For this we consider a spherical Gaussian surface of radius \(r < R\) which passes through point ‘\(P\)’ as shown in the figure below.
FIGURE 07
Spherical Gaussian surface inside a uniformly charged spherical shell.
Because the Gaussian surface enclose no charge, therefore ‘q = 0’,
\[\oint\vec{E}.d\vec{A}= 0 \]
\[\ointE dA \cos {0°}= 0 ∴\vec{E}∥d\vec{A}\]
\[E\oint dA= 0 ∴E is constant\]
As \(dA≠0\), therefore
\[E= 0 \]
So the electric field does not exist inside a uniform shell of charge. So the test charge placed inside the charged shell would experience no force.
CHAPTER 29
29.5.4 Electric Field due to Spherical Charge
Question.Find out the expression of electric field intensity outside solid sphere of charge.
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FIGURE 08
Gaussian surface outside a uniformly charged solid sphere.Ans. Consider a spherical distribution of charge of radius ‘\(R\)’ with the uniform volume charge density ‘\(\rho \)’. We want to find the electric field at point ‘\(P\)’ at a distance \(r > R\) from the center of charged sphere. For we consider a spherical Gaussian surface which passes through point ‘\(P\)’ as shown in the figure below.
As the electric field is radial, so the electric lines of force leave the Gaussian surface normally at all points. Therefore, the electric field intensity \(\vec{E}\) and surface area element \(d\vec{A}\)are in same direction i.e., \(\theta =0°\).
\[\ointE dA= \frac{q}{{\epsilon}_{0}}\]
\[E \ointdA= \frac{q}{{\epsilon}_{0}}∴E is constant\]
Thus for all points outside the spherical charge distribution, the electric field has the same value as if the charge is concentrated at the center of sphere.
Q. Derive the expression of electric field intensity inside solid sphere of charge.
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Ans. Consider a spherical distribution of charge of radius ‘\(R\)’ with the uniform volume charge density ‘\(\rho \)’. The total charge in this uniform charge distribution is
We want to find the electric field at point ‘\(P\)’ at a distance \(r < R\) from the center of charged sphere. For this we consider a spherical Gaussian surface of which passes through point \(P\)as shown in the figure below.
FIGURE 09
Gaussian surface inside a uniformly charged solid sphere.Let the Gaussian surface encloses the charge \(q' < q\) given by
Variation of Electric Field Intensity as a function of distance for Volume charge distribution
The graphical representation of the dependence of electric field strength on the radial distance ‘\(r\)’ from the center of this charge distribution is shown in the figure:
FIGURE 10
Variation of electric field intensity with radial distance for a uniformly charged solid sphere.Electric field intensity, inside solid sphere of charge, is directly proportional to distance as described by formula:
The solid sphere of charge behaves as a point charge for all the points outside the solid sphere of charge i.e., electric field intensity is inversely proportional to the square of the distance from center of sphere of charge.
Problem 21. The drum of a photocopying machine has the length of 42 cm and diameter of 12 cm having surface charge density equal to \(2\times {10}^{-6}C/m^{2}\). What is the total charge on the drum? (b) The manufacturer wants to produce desktop version of machine. This requires reducing the length of drum to 28 cm and diameter of 8 cm. the electric field must remain unchanged. Calculate the charge of new drum.
Spherical Gaussian surface surrounding a point charge.Coulomb’s law can be deduced from Gauss’s law under certain symmetry consideration. Consider positive point charge ‘\(q\)’. In order to apply the Gauss’s law, we assume a spherical Gaussian surface as shown in the figure below.
This equation gives the magnitude of electric field intensity \(E\) at any point which is at the distance ‘\(r\)’ from an isolated point charge ‘\(q\)’.
From the definition of electric field intensity, we know that
\[F = q_{0 }E\]
Where \(q_{0 }\)is the point charge placed at a point at which the value of electric field intensity has to be determined. Therefore
29.7 Prove that “An excess charge added to the isolated conductor moves entirely to its outer surface. None of the excess charge is found within the body of conductor”.
Consider an isolated conductor (lump of copper) is hanging from a silk thread and carrying a net positive charge ‘q’ as shown in the figure below. The Gaussian surface lies inside the actual surface of the conductor.
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FIGURE 12
Gaussian surface inside an isolated charged conductor.
Under the equilibrium conditions, the electric field inside the conductor must be zero. If it were not so, the field would exert the force on conduction electrons and the internal currents would be setup.
But there is no experimental evidence of such internal currents in isolated conductors. And if the extra charge is added to the surface, it redistribute itself on the surface in such a way that the electric field inside the conductor vanish.
If E is zero everywhere inside the conductor, it must be zero at all the points of Gaussian surface. This means that the flux through Gaussian surface must be zero. Gauss’s law (\({\Phi}_{e}=\frac{q}{{\epsilon}_{0}}\)) then tells that the net charge inside the Gaussian surface must also be zero.
If the added charge is not inside the Gaussian surface, it can only be outside that surface. And the added charge must lie on the actual outer surface of the conductor.
CHAPTER 29
29.8 Prove that the formation of cavity by cutting a natural material from the conductor does not change the distribution charge or pattern of electric field.
FIGURE 13
Gaussian surface around a cavity inside a charged conductor.Consider an isolated conductor hanging from a silk thread carrying a net positive charge ‘q’. Suppose a cavity is produced inside the conductor as shown in the figure below.
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We draw a Gaussian surface around the cavity inside the conductor. Because E is zero inside the conductor, there is no flux though the Gaussian surface. So by Gauss’s law, there is no net charge inside the Gaussian surface. So the total charge remains on the outer surface of conductor.
CHAPTER 29
29.9 The External Electric Field
FIGURE 14
Cylindrical Gaussian surface used to determine the external electric field of a charged conductor.The electric field outside a charged isolated conductor can be find out by Gauss’s law by considering the cylindrical Gaussian surface as shown in the figure below.
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The flux through the interior end cap is zero, because E = 0 for all interior points of conductor. The flux through the cylindrical walls is also zero because the lines of E are parallel to the surface. But the flux through the outer cap will not be zero.
This showed that the electric field intensity at any point is doubled than the value of E for an infinite sheet of charge.
Sample problem 3. The electric field just above the surface of the charged drum of a photo copying machine has a magnitude E of \(2.3\times {10}^{5}N/C\). What is the surface charge density on the drum if it is a conductor?
Sample problem 4. The magnitude of the average electric field normally present in the earth atmosphere just above the surface of the earth is \(150 N/C\) directed downward. What is the total net charge carried by the earth? Assume the earth to be conductor.
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