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1.1 Electric Charge
Electrical Charge and Coulomb’s Law
Electromagnetic forces are responsible for the structure of atoms and for the binding of atoms in molecules and solids. Many properties of materials that we have studied so far are electromagnetic in their nature. Such as the elasticity of solids and the surface tension of liquids. The spring force, friction, and the normal force all originate with the electromagnetic force between atoms.
Among the examples of electromagnetism that we shall study are the force between electric charges, such as occurs between an electron and the nucleus in an atom; the motion of a charged body subject to an external electric force. Such as an electron in an oscilloscope beam; the flow of electric charges through circuits and the behavior of circuit elements; the force between permanent magnets and the properties of magnetic materials; and electromagnetic radiation, which ultimately leads to the study of optics, the nature
and propagation of light.
In this chapter, we begin with a discussion of electric charge, some properties of charged bodies, and the fundamental electric force between two charged bodies.
Electric Charge and Electrical Forces
A body is said to be electrical neutral if it contains equal number of +ve and –ve charges. When two bodies are rubbed together, their neutrality is distributed due to transfer of electrons from one body to the other. The body which gives electrons becomes electrically positive and the body which gains electrons becomes negative.
“Charges of the same signs repel each other and Charges of the oppositely sign attract each other.”
These attractive and repulsive forces among the charges are called electrical forces.
Q # 1. Explain what is meant by the term “a neutral atom.’’ Explain what “a negatively charged atom’’ means.
Ans. A neutral atom is one that has no net charge. This means that it has the same number of electrons orbiting the nucleus as it has protons in the nucleus. A negatively charged atom has one or more excess electrons.
Q # 2. Explain from an atomic viewpoint why charge is usually transferred by electrons.
Ans. Electrons are less massive and more mobile than protons. Also, they are more easily detached from atoms than protons.
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1.2 Coulomb’s Law
Q # 3. Would life be different if the electrons were positively charged and the protons were negatively charged? Does the choice of signs have any bearing on physical and chemical interactions? Explain.
Ans. No. Life would be no different if electrons were + charged and protons were – charged. Opposite charges would still attract, and like charges would repel. The naming of + and – charge is merely a convention.
(Definition) Point Charges:
The charge bodies whose sizes are much smaller than the distance between them are called point charges.
Charles Augustin Coulomb (l736- 1806) measured electrical attractions and repulsions quantitatively
and deduced the law that governs them. His apparatus, shown in Fig. consist of spheres A and B.
If A and B are charged, the electric force on A tends to twist the suspension fiber. The angle is then a relative measure of the electric force acting on charge
A. Experiments due to Coulomb and his contemporaries showed that
Statement
The magnitude of electrical force between two point charges is directly proportional to the product of magnitude of charges and inversely proportional to the square of the distance between their centers and the force acts along the line connecting the charges.
Mathematical Form
Suppose two point charges ?1 and ?2 are separated by distance ?. According to the Coulomb’s law, the electrical force between two charges is: ?∝?1?2 −−−−−(1) ?∝1?2 −−−−−−(2)
Combining (1) and (2) we have, ?∝?1?2?2 ??,?=??1?2?2
Where k is called Coulomb’s constant. Its value depends upon the system of units and medium between the charges. For free space and in system international ‘k’ is expressed as: ?=14??0
Where ?0 is the permittivity of free space and its value in SI unit is: 8.854×10−12 ?2?−1?−2
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Example 1. What must be the distance between point charge ??=??.? ?? and point charge ??=−??.? ?? for the attractive electrical force between them to have a magnitude of 5.66 N.
Solution:
?1=26.3 ??=26.3×10−6 ?
?2=−47.1 ??=−47.1×10−6 ?
?=5.66 ?
From Coulomb’s law:
?=??1?2?2
Or,?=√??1?2?=√(9×109)×(26.3×10−6)×(47.1×10−6)5.66=1.40 ?
Example 2. In the radioactive decay of ????, the center of the emerging ???? particle is at a certain distance ??×??−?? ? from the center of residual ?? ???nucleus at that instant. (a) What is the force on helium atom and (b) what is its acceleration?
Solution:
? 238→?? 24+?ℎ90234 ?=12×10−15 ? ?=? ?=? ?=??1?2?2=9×109×(2?)×(90?)(12×10−15)2=288 ?
Mass of helium atom ?=4 ???=4×1.67×20−27 ??=6.68×10−27 ??
From Newton’s second law of motion: ?=??⇒?=??=2886.68×10−27=4.311×1028 ??−2
So far, we have considered only the magnitude of the force between two charges determined according
to Coulomb’s law. Force, being a vector, has directional properties as well. In case of coulomb’s law, the direction of the force is determined by the relative sign of the two
electric charges.
Suppose we have the two point charges ′?1′ and ′?2′ separated by a distance ?12. For the moment, we assume the two charges to have the same sign, so that they repel each other. Let us consider the force exerted on ?1 by ?2 is denoted by ?⃗⃗⃗ 12 .The position
1.3 Vector Form of Coulomb’s Law
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vector that locates particle 1 relative to particle 2 is ? 12 ; that is, if we were to define the origin of our coordinate system at the location of ?2 , then ? 12 would be the position vector of ?1.
If the charges have the same sign, then the force is repulsive
and ? 12 must be parallel to ? 12, as shown in figure a. If the charges are of opposite sign, as in figure b, then ? 12 is attractive and anti-parallel to ? 12. In either case we can represent the force as: ? 12=??1?2?122 ?̂12
Question: Show that Coulomb force is a mutual force
Ans: Coulomb’s force is a mutual force, it means that if a charge ′?1′ exerts a force on charge ′?2′ , then ′?2′ also exerts an equal and opposite force on ′?1′.
If ?̂12 represents the direction of force exerted on charge ?1 by ?2 and ?̂21 is the unit vector which represent the direction of force on charge ?2 by ?1, then,
? 21=14??0?1?2?2 ?̂21−−−−−−−(1) ? 12=14??0?1?2?2 ?̂12−−−−−−−(2)
As ?̂21=−?̂12 , so the eqn. (1) becomes
? 21=14??0?1?2?2 (−?̂12) =−14??0?1?2?2 ?̂12
By eqn (2) ? 21=? 12
This expression shows that Coulomb force is a mutual force.
1.3.1 Significance of Vector Form of Coulomb’s Law
Vector form of coulomb’s law has the critical importance, when there is an assembly of point charges. In this case, the resultant force on any one of the charges is the vector sum of the forces due to each of the other forces. This is called principle of superposition.
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1.3.2 Coulomb force due to many point charges
Let ?1,?2,?3………?? are the ‘n’ point charges as shown in the figure. We want to find out electrical force on charge
?1 by the assembly of n point charges. The point charges ?2,?3,?4………?? are at the distances ?12,?13,?14………?1? from charge ?1 respectively.
Now if ? 12,? 13,? 14………? 1? be the electrostatic force
on charge ?1 due to the point charges ?2,?3,?4………?? respectively. Thus, the total electrostatic force ? 1 on charge ?1is given by; ? 1=? 12+? 13+? 14+⋯……+? 1?−−−−−−(1)
Where,
? 12= Force on charge ′?1′ exerted by ′?2′ =14??0?1?2?122 ?̂12
? 13= Force on charge ′?1′ exerted by ′?3′ =14??0?1?2?132 ?̂13
… … … …
… … … …
? 1?= Force on charge ′?1′ exerted by ′??′ =14??0?1???1?2 ?̂1?
Putting values in equation (1) we get, ? 1=14??0?1?2?122 ?̂12+14??0?1?2?132 ?̂13+⋯……+14??0?1???1?2 ?̂1? =?14??0(?2?122 ?̂12+?2?132 ?̂13+⋯……+???1?2 ?̂1?) =?14??0 Σ???1?2 ?̂1???=1
This expression gives the electrical force between on a point charge due to many point charges.
When the two bodies are rubbed together, transfer of electrons from one body to the other takes place and they are said to be electrified. The magnitude of charge q that can be detected and measured on any object is given by
?=?? ---------------------- (1)
where ?=0,±1,±2,……… and ? is the elementary unit of charge called on unit charge, has the experimentally determined value
?=1.6×10−19 ?
1.4 Quantization of Charges
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When a physical quantity is discrete values, it is called quantized quantity.
Equation (1) shows that charge is also a quantized quantity like matter, energy, angular momentum etc. It means that we can find a body that can have a charge of 10? or −5? but it is not possible to find a body with fractional charge such as +3.57? or −2.35?.
Example-3. The electrostatic force between identical ions that are separated by a distance of ?×??−?? ? is ?.?×??−? ? (a) Find the charge on each ion? (b) How many electrons are missing from each ion?
Solution: ?=3.7×10−9 ?
?=5×10−10 ?
(a) ?1=?2=?=?
?=??1?2?2=??2?2
??,?2=??2?=3.7×10−9×(5×10−10)29×109=10.27×10−38
?=3.20×10−19 ?
(b) ?= ?
?=??
?=??=3.20×10−191.6×10−19=2
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THE ELECTRIC FIELD
Electric charges interact with each other over vast distances. Electrons or ionized atoms at the furthest reaches of the known universe can exert forces that cause electrons to move on the Earth. How can we explain these interactions? We do so in terms of electric field- the distant charge set up an electric field, which exist throughout the space between the Earth and is the origin of the field.
In this chapter we consider only the static electric field due to charges at rest.
To describe the mechanism by which one charge particle exert the force the force on other charge particles,
Michael Faraday introduced the concept of electric field.
Electric Field
The region or space around a charge in which it can exert the force of attraction or repulsion on other charged bodies is called electric field.
Electric Field Intensity
The electrostatic force on unit positive charge at a specific field point is called the electric field intensity. In order to find out electric field intensity, a test charge ?0 is placed in the electric field at a field point. The electric field intensity ?⃗ is expressed as, ?⃗ =? ?0
Where ? is the electrostatic force on test charge ?0.
The test charge ?0 should be very small, so that it cannot disturb the field produced by source charge
?.Therefore the electric field intensity can be written as, ?⃗ =lim?0→0? ?0
Electric Field Lines or Lines of Force
A visual representation of the electric field can be obtained in terms of electric field lines. Electric field lines can be thought of a map that provides information about the direction and strength of the electric field at various places. As electric field line provides the information about the electric force exerted on a charge, the lines are commonly called “Lines of Force”.
Properties of Electric Field Lines
i) Electric field lines originate from positive charges and end on negative charges.
ii) The tangent to a field line at any point gives the direction of the electric field intensity at that point.
iii) The lines are closer where the field is strong, the lines are farther apart where the field is weak.
iv) No two lines cross each other.
2.1 Electric Field Intensity
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Conceptual Q # 1. When defining the electric field, why is it necessary to specify that the magnitude of the test charge be very small?
Ans. So the electric field created by the test charge does not distort the electric field you are trying to measure, by moving the charges that create it.
Conceptual Q # 2. An object with negative charge is placed in a region of space where the electric field is directed vertically upward. What is the direction of the electric force exerted on this charge?
Ans. Vertically downward.
Conceptual Q # 3. Is it possible for an electric field to exist in empty space? Explain.
Ans. An electric field once established by a positive or negative charge extends in all directions from the charge. Thus, it can exist in empty space if that is what surrounds the charge.
Conceptual Q # 4. Explain why electric field lines never cross.
Ans. The direction of the electric field is the direction in which a positive test charge would feel a force when placed in the field. A charge will not experience two electrical forces at the same time, but the vector sum of the two. If electric field lines crossed, then a test charge placed at the point at which they cross would feel a force in two directions, which is not possible.
Consider a test charge ?0 placed at point P in the electric field of a point charge ? at a distance ? apart.
We want to find out electric field intensity at point ? due to a point charge ?.
The electrostatic force ? between ? and ?0 can be find out by using expression,
?=14??0??0?2
The electric field intensity ? due to a point charge ? can be obtained by putting the value of electrostatic force in expression of electric field intensity:
?=(14??0??0?2)?0
?=14??0??2
This expression gives the magnitude of electric field intensity due to a point charge ?. In vector form, the electric field intensity ?⃗ will be: ?⃗ =14??0??2 ?̂
2.2 Electric Field Intensity Due To a Point Charge
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Where ?̂ is the unit vector which gives the direction of electric field intensity.
Conceptual Q # 5. Explain what happens to the magnitude of the electric field created by a point charge as r approaches zero.
Ans. The electric field around a point charge approaches infinity as r approaches zero.
Example-4. In an ionized helium atom (a helium atom in which one of the two electrons has been removed) the electron and nucleus are separated by a distance of 26.5 pm. What electric field due to the nucleus at the location of the electron.
Solution:
Total charge of helium nucleus 2?=2×1.6×10−19=3.2×10−19 ?
Distance ?=26.5 ??=26.5×10−12 ? ?=???2=9×109×3.2×10−19(26.5×10−12)2=4.1×1012 ??−1
Example-5. Two equal and opposite charges of magnitude ?.??×??−? ? are held 15.2 cm apart. What is the direction and magnitude of E at mid-point between the charges? What is the force act on an electron placed here?
Solution: ?1=1.88×10−7? ?2=1.88×10−7?
Distance, ?=15.2 ??=15.2×10−2 ?
Mid-point, ?=?2=7.6×10−2 ?
Total electric field ?=|?+|+|?−|= ?
Force on an electron placed at the same point ?= ? |?+|=??1?2=9×109×1.88×10−7(7.6×10−2)2=1.28×106?? |?−|=??2?2=9×109×1.88×10−7(7.6×10−2)2=1.28×106??
Total electric field ?=|?+|+|?−|=2.56×106??
Now, Force on an electron placed at the same point ?=??=−1.6×10−19×2.56×106?? =−4.096×10−13?
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GAUSS’S LAW
Coulomb’s law can always be used to calculate the electric field intensity 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 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.
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 ??.
Consider a surface ′?′ placed in a uniform electric field of intensity ?⃗ . Let ? be the area of the surface. The component of ?⃗ normal to the area ? is ????? as shown in the figure below.
The electric flux through the surface ? is given by;
??=?(?????)
??=??????
??=?⃗ .?
Thus, the electric flux is the scalar product of electric field intensity and the vector area. The SI unit of
the electric flux is ??2?
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.
We divide the surface into n number of small patches having area Δ?1,Δ?2,Δ?3,………,Δ??.
Let ?⃗ 1,?⃗ 2,?⃗ 3,………,?⃗ ? are the electric field intensities which makes angle ?1,?2,?3,………,??with the normal to the area elements
3.1 Electric Flux
3.2 Electric Flux through an Irregular Shaped Object
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Δ?1,Δ?2,Δ?3,………,Δ?? respectively. If ?1,?2,?3,………,?? be the electric flux through Δ?1,Δ?2,Δ?3,………,Δ??, then the total electric flux ?? will be: ??=?1+?2+?3+⋯……+??
⇒??=?1(Δ?1????1)+?2(Δ?2????2)+⋯……+??(Δ???????) ⇒??=?1Δ?1????1+?2Δ?2????2+⋯……+?????????
⇒??=?⃗ 1.Δ? 1+?⃗ 2.Δ? 2+⋯……+?⃗ ?.Δ? ?.
Where Δ? 1,Δ? 2,………Δ? ? are the vector areas corresponding to area elements Δ?1,Δ?2,………Δ?? respectively
??=Σ?⃗ ?.Δ? ???=1
??=∫?⃗ .?? ?
By convention, the outward flux is taken as positive and inward flux is taken as negative
Example-6. Consider a hypothetical closed cylinder of radius ? immersed in a uniform electric field ?⃗⃗ , the
cylinder axis being parallel to the field What is ?? for this closed surface?
Solution:
The flux ?? can be written as the sum of three terms, an integral over (a) the left cylinder cap, (b) the cylindrical surface, and (c) the right cap. ??=∮?⃗ .?? =∫?⃗ .?? +∫?⃗ .?? ?+∫?⃗ .?? ? ?
For the left cap, the angle ? for all points is 180°, ?⃗ has a constant value, and the vectors ?? are all parallel. Thus, ∫?⃗ .?? =∫??????180°=−?∫??=−?? ?
Similarly, for the right cap, ∫?⃗ .?? =∫??????0°=?∫??=?? ?
The angle ? for all points being 0° here.
Finally, for the cylinder wall, ∫?⃗ .?? =0?
When ?→∞,?? Δ?→0, then the sigma is replaced by the surface integral i.e,
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Because, ?=90°; hence, ?⃗ .?? =0 for all points on the cylindrical surface. Thus the total flux is, ??=−??+0+??=0
This result is expected, because the filed lines that enter at the left goes out from the right end.
Statement:
The total electric flux through any close surface is 1?0 times the total charge enclosed by the surface.
??=? ?0−−−−(1)
Where ? is the net charge enclosed by the surface. Also,
??=∮?⃗ .?? −−−(2)
Comparing (1) and (2) we have,
∮?⃗ .?? =??0
Thus we can describe the Gauss’s law as
The surface normal integral of electric field intensity is equal to 1?0 times the total charge
enclosed by the surface.
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.
Consider a section of infinite line of charge having uniform linear charge
3.4 Applications of Gauss’s law
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:
3.3 Gauss’s Law
3.4.1 Electric Field due to Infinite Line of Charge
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∫??=2??ℎ?3
So, equation (1) becomes:
??=?(2??ℎ)=2??ℎ?−−−(2)
By, Gauss’s law
??=??0−−−−(3)
As the line of charge has constant linear charge density ?, therefore: ?=?/ℎ ⇒?=?ℎ
So, the equation (3) becomes:
??=?ℎ?0−−−(4)
density ‘?’ as shown in the figure below.
We want to find out electric field intensity at any point ′?′ which is at distance ‘?’ from the wire. For this we consider cylindrical Gaussian surface which
passes through point ′?′.The electric flux passing through the cylinder is given as
??= ?⃗ .??
The surface ‘?’ of the cylinder consist of three parts i.e., ?1,?2,?3, where
?1= Area of top cross section of cylindrical Gaussian surface
?2= Area of bottom cross section of cylindrical Gaussian surface
?3= Area of curved part of Gaussian surface
Thus ??=∫?⃗ .?? +∫?⃗ .?? +∫?⃗ .?? ?3?2?1
Now
∫?⃗ .?? =∫?⃗ .?? =0?2?1
And, ∫?⃗ .?? =∫??????0°=∫????3?3?3
Therefore
??=?∫??−−−−−(1)?3
For cylindrical symmetry,
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Comparing Eq. (2) and (4), we get: 2??ℎ ?=?ℎ?0
?=?2???0
If ?̂ gives the direction of electric field intensity then, ?⃗ =?2???0 ?̂
This expression gives the electric field intensity due to infinite line of charge.
Example-7. An infinite line of charge produces a field of ?.??×??? ?/? at a distance of 1.96 m. Calculate the linear charge density?
Solution:
?=?.??×?????
?=?.?? ?
?=?
As, ?=??????
?=?????.?
?=?×?.??×?.??×??−??×?.??×?.??×???=?.??×??−???
Example-8. A plastic rod whose length is 220 cm and whose radius is 3.6 mm carries a negative charge q of magnitude ?.?×??−? ? spread uniformly over its surface. What is the electric field near the midpoint of the rod at a point on its surface?
Solution: ?=220 ??=2.2 ? ?=3.6×10−3 ? ?=−3.8×10−7 ? ?=?
As Electric field intensity due to infinite line of charge is: ?=?2??0?−−−−−(1)
?=??=−3.8×10−72.2=−1.73×10−7??
Now, ?=?2??0?= −1.73×10−72×3.14×8.85×10−12×3.6×10−3=−8.6×105??
Proof: Consider a thin spherical shell of radius ‘?’ which have the charge
‘?’ with constant surface charge density ‘?’.
Question: Show that the uniform spherical shell of charge behaves, for all external points, as if all its
charge were concentrated at its center.
Consider a point ‘?’ 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 ?>? which passes through point
′?′ as shown in the figure below.
According to Gauss’s law, ∮?⃗ .?? =??0 ∮??????0°=??0 [?⃗ ∥?? ] ?∮??=??0 [ ? ?? ????????] ?(4??2)=??0 ?=14??0??2
∮??????0°=0 ?∮??=0
As ??≠0, therefore
3.4.2 Electric Field due to Spherical Shell of Charge
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.
Consider a point ‘?’ inside the shell. We want to fine out electric field
intensity ‘?’ at point ‘?’ due to this symmetrical charge distribution. For this we consider a spherical Gaussian surface of radius ?<? which passes through point ‘?’ as shown in the figure below.
According to Gauss’s law,
∮?⃗ .?? =??0
Because the Gaussian surface enclose no charge, therefore ‘q = 0’, ∮?⃗ .?? =0
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3.4.3 Deduction of Coulomb’s Law from Gauss’s Law
?=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.
Coulomb’s law can be deduced from Gauss’s law under certain symmetry consideration. Consider positive point charge ‘?’. In order to apply the Gauss’s law, we assume a spherical Gaussian surface as shown in the figure below.
Considering the integral form of Gauss’s law, ∮?⃗ .?? =??0
Because the both vectors ?⃗ and ?? are directed radially outward, so ∮??????0°=??0
As E is constant for all the points on the spherical Gaussian surface,
?∮??=??0 ?(4??2)=??0 ?=14??0??2
This equation gives the magnitude of electric field intensity ? at any point which is at the distance ‘?’ from an isolated point charge ‘?’.
From the definition of electric field intensity, we know that
?=?0?
Where ?0 is the point charge placed at a point at which the value of electric field intensity has to be determined Therefore ?=14??0??0?2
This is the mathematical form of Coulomb’s law.
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ELECTRIC POTENTIAL
The energy approach in the study of dynamics of the particles can yield not only the simplification but also new insights. One advantage of energy method is that, although force is a vector, energy is a scalar. In problems involving vector forces and fields, calculations involving sums and integrals are often complicated ??→??0=−∫?⃗ .?? ??
as, ??→??0=Δ?=??−??
Therefore, the electrical potential difference between two points in an electrical field will be,
In this chapter, we introduce the energy method to the study of electrostatics.
Potential difference ′Δ?′ between two point is defined as “the amount of work done ′Δ?′ per unit charge ′?0′ in moving it from one point to the other against the electric field and by keeping the system in equilibrium”. Mathematically Δ?=Δ??0
Suppose a unit positive test charge ′?0′ is moved from one point ′?′ to the point ′?′ in the electric
field ?⃗ of a large positive charge ′?′ as shown in figure below:
The work done in moving ′?0′ from point ′?′ to the point ′?′ against the electric field ?⃗ is ??→?=∫? .?? ??
The electrical force of magnitude ? =−?0?⃗ must have to supplied in order to move ′?0′ against the electric field. Therefore ??→?=∫ −?0?⃗ .?? ?? ??→?=−?0∫?⃗ .?? ??
4.1 Potential Difference
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??−??=−∫?⃗ .?? ??
Absolute electric potential at a point is defined as “the amount of work done per unit charge in
moving it from infinity to a specific field point against the electric field and by keeping the system in equilibrium”.
To find the absolute potential, the reference point is selected at which potential is zero. This point is situated at infinity i.e., out of the electric field. Thus, in equation (1) ??=?(∞)=0
Thus, ??−0=−∫?⃗ .?? ?∞
If the distance from the point ′?′ to the charge ′?′ is ′?′, then in general ?(?)=−∫?⃗ .?? ?∞
The potential difference between two points is the amount of work done per unit charge ′?0′ in
moving it from one point to the other against the electric field ′?⃗ ′. Mathematically, it is described as: ??−??=−∫?⃗ .?? ????
But the electric field intensity due to point charge: ?⃗ =14??0??2 ?̂
So, ??−??=−∫14??0??2 ?̂.?? ???? ??−??=−?4??0∫1?2 ?̂.?? ????
As, ?⃗ is directed radially outward, therefore ?̂∥??
hence, ?̂.?? =|?̂||?? |???0°=(1)(??)(1)=??
4.2 Absolute Electrical Potential at a Point
4.3 Expression for the Electric Potential Difference due to a Point Charge
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So, ??−??=−?4??0∫???2 ???? ??−??=−?4??0 |−1?|????=?4??0 |1?|???? ??−??=?4??0 [1??−1??]
This is the expression for the potential difference between two points ′?′??? ′?′.
The electric potential at any point is the amount of work done per unit charge in moving a unit
positive charge (test charge) from infinity to that point, against the electric field. If the point ′?′ is at infinity then ??=?(∞)=0, ??=∞
Putting this value in equation in the expression of electric potential difference due to point charge, we get: ??−0=?4??0 [1??−1∞] ??=?4??0 [1??]
In general, the electric potential at point due to a point charge ′?′ is ?=14??0??
Example-9. Two protons in the nucleus of ???? are ? ?? apart. What potential energy associated with the electric force that acts between them?
Solution: ?=6 ??=6×10−15 ? Δ?= ? ?1=?2=1?=1.6×10−19 ?
As, ΔV=kqr=9×109×1.6×10−196×10−15=2.34×105 V
Now, Δ?=?1.Δ?=1.6×10−19×2.34×105=3.744×10−14 ?
Δ?=3.744×10−141.6×10−19=2.34×105 ??
Example-10. What is the electric potential at the surface of the gold nucleus. The radius of the gold nucleus is ?×??−?? ? and atomic number is 79.
Solution: ?=7×10−15 ? ?=79 ?=79?=79×1.6×10−19 ? ?= ?
4.4 Expression for the Absolute Electric Potential due to a Point Charge
20
As, ?=???=9×109×79×1.6×10−197×10−15=1.6×107 ?