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Electrostatics
Electric charge
Electrically charged objects have several important characteristics:
A conductor is a material through which electric charges can easily flow. An insulator is a material through which electric charges do not move easily, if at all.
An electroscope is a simple device used to indicate the existence of charge. As shown in Figure 1, the electroscope consists of a conducting knob and attached lightweight conducting leaves—commonly made of gold foil or aluminum foil.
When a charged object touches the knob, the like charges repel and force the leaves apart. The electroscope will indicate the presence of charge but does not directly indicate whether the charge is positive or negative.
Figure 1
An electroscope reports the presence of charge.
A large charge near a neutral electroscope can make the leaves move apart. The electroscope is made of conducting material, so the positive charges are attracted to the knob by the nearby (but not touching) negatively charged rod. The leaves are left with a negative charge and therefore deflect. When the negative rod is removed, the leaves will fall.
Now, consider touching the electroscope knob with a finger while the charged rod is nearby. The electrons will be repulsed and flow out of the electroscope through the hand. If the hand is removed while the charged rod is still close, the electroscope will retain a charge. This method of charging is called charging by induction
Electrostatics, is the study of stationary electric charges. A rod of plastic rubbed with fur or a rod of glass rubbed with silk will attract small pieces of paper and is said to be electrically charged. The charge on plastic rubbed with fur is defined as negative, and the charge on glass rubbed with silk is defined as positive.
When an object is rubbed with a charged rod, the object shares the charge so that both have a charge of the same sign. In contrast, charging by induction gives an object the charge opposite that of the charged rod.
Even though the charges are not free to travel throughout the material, insulators can be charged by induction. A large charge nearby—not touching—will induce an opposite charge on the surface of the insulator. As shown in Figure 3, the negative and positive charges of the molecules are displaced slightly. This realignment of charges in the insulator produces an effective induced charge.
Coulomb’s law
Coulomb’s law gives the magnitude of the electrostatic force ( F) between two charges:
where q 1 and q 2 are the charges, r is the distance between them, and k is the proportionality constant. The SI unit for charge is the coulomb. If the charge is in coulombs and the separation in meters, the following approximate value for k will give the force in newtons: k= 9.0 × 10 9 N · m 2/C 2. The direction of the electrostatic force depends upon the signs of the charges. Like charges repel, and unlike charges attract.
Coulomb’s law can also be expressed in terms of another constant (ε 0), known as the permittivity of free space:
When the permittivity constant is used, Coulomb’s law is
The most fundamental electric charge is the charge of one proton or one electron. This value (e) is e = 1.602 × 10 −19 coulombs. It takes about 6.24 × 10 18 excess electrons to equal the charge of one coulomb; thus, it is a very large static charge.
For a system of charges, the forces between each set of charges must be found; then, the net force on a given charge is the vector sum of these forces. The following problem illustrates this procedure.
Example 1: Consider equal charges of Q whose value in coulombs is not known. The force between two of these charges at distance X is F. In Figure , three charges (3 Q) are placed at point A, which is a distance X from point B. One charge (Q) is placed at point B, which is X/2 distance from point C, which has one charge. What is the net force on the charge at point B?
Solution: This problem can be solved through proportional reasoning. The force of 3 Q on the one charge at B will be 3 F. Because the single charge is one‐half X from B, the force will be four times greater than at a distance X, that is, 4 F. The forces of 3 F and 4 F are at right angles, and therefore, the resultant force is 5 F, or
The direction is found from the tangent: θ −1 = tan 4/3 = 53°
Electric fields and lines of force
When a small positive test charge is brought near a large positive charge, it experiences a force directed away from the large charge. If the test charge is far from the large charge, the electrostatic force given by Coulomb’s law is smaller than when it is near. This data of direction and magnitude of an electrostatic force, due to a fixed charge or set of fixed charges, constitutes an electrostatic field. The electric field is defined as the force per unit charge exerted on a small positive test charge (q 0) placed at that point. Mathematically,
Note that both the force and electric field are vector quantities. The test charge is required to be small so that the field of the test charge does not affect the field of the set charges being examined. The SI unit for electric field is newtons per coulomb (N/C).
Figure 5 is a pictorial representation of the electric fields surrounding a positive charge and a negative charge. These lines are called field lines or lines of force.
The rules for drawing electric field lines for any static configuration of charges are
Electric flux
Electric flux is defined as the number of field lines that pass through a given surface. In Figure , lines of electric flux emerging from a point charge pass through an imaginary spherical surface with the charge at its center
This definition can be expressed as follows: Φ = ∑E · A, where Φ (the Greek letter phi) is the electric flux, E is the electric field, and A is area perpendicular to the field lines. Electric flux is measured in N · m 2 / C 2 and is a scalar quantity. If the surface under consideration is not perpendicular to the field lines, then the expression is Φ = ∑ EAcos θ.
In general terms, flux is the closed integral of the dot product of the electric field vector and the vector ΔA. The direction of ΔA is the outward drawn normal to the imaginary surface. Mathematically, Φ = ΦE · dA. The accepted convention is that flux lines are positive if leaving a surface and negative if entering a surface.
Gauss’s law
or in integral form,
where θ is the angle between the direction of E and the outward direction of normal to the surface and ε is the permittivity constant.
Consider the calculation of the electric field due to a point charge. Figure shows the point charge, the direction of its field, and a gaussian surface. Because the electric field is perpendicular to the gaussian surface and directed outward, θ is 90 degrees, and cos θ = 1. Gauss’s law is
Substitute in the area of a sphere, and the left side reduces to
or
which is the same expression obtained from Coulomb’s law and the definition of electric field in terms of force.
The derivation of the expression for the field due to a thin conducting shell of charge follows. Figure 8 shows the electric fields for (a) a shell of radius ( R), (b) the gaussian surface for outside the shell, and (c) the gaussian surface for inside the shell (c) of radius ( r).
When outside the shell of charge, as in Figure (a), the left side of Gauss’s equation reduces to the following expression for the same reasons given for a point charge:
Therefore,
Thus, the electric field outside a sphere of charge is the same as if the same amount of charge were concentrated in a point located at the center of the sphere.
The gaussian surface inside the sphere encloses no charge, and therefore, there is no electric field inside the uniformly charged spherical shell. The same proof holds within a solid conductor because all the charge of the conductor resides on the surface. Because the electric field inside even an irregularly shaped conductor is zero, the charge will not be evenly distributed over an irregular shape. The charge will tend to accumulate on protruding points on the outside of the conductor.
Potential difference and equipotential surfaces
In the preceding examples, the charge distributions were spherical, and so the gaussian surface was a sphere. When finding the electric field of either a sheet of charge or a line of charge, a cylinder is the correct gaussian surface to use.
Example 2: Find the electric field for a nonconducting infinite sheet of charge. The electric field is directed outward from the sheet on both sides. The charge per unit area is σ (the Greek letter sigma). See Figure for the electric field and gaussian surface.
The area of the closed cylindrical gaussian surface will be the sum of the areas of the left end, the right end, and the wall; therefore, ∑ EAcos θ = ( EA cos θ) left end + ( EA cos θ) right end + ( EA cos θ) wall. The electric field is parallel to the wall, which is at right angles to the outward normal of the wall area; thus, the last term on the right is zero. At each end, E is in the same direction as the outward normal, so ( EA cos θ) left end + ( EA cos θ) right end = 2 EA, where A is the area of the end of the gaussian cylinder. The total charge inside the gaussian surface is the product of the charge per unit area and the area; so
and
Note that the magnitude of the electric field does not depend upon the distance from the plate. The electric field is uniform. In the practical case of finite plates of charge, the electric field is uniform relatively close to the charged plate.
The resultant electric field of two parallel plates is double that of one sheet with the same charge:
or
where q is the charge on each plate, and A is the area of each plate. If the plates have opposite charges, the electric field will exist between the plates and be zero outside the plates. If the charges are of equal sign, the electric field will be zero between the plates and be expressed by the above equation outside the plates. These results can be derived by Gauss’s law.
Electrostatic potential and equipotential surfaces
Imagine moving a small test charge q′ from point A to point B in the uniform field between parallel plates. The work done in transferring the charge equals the product of the force on the test charge and the parallel component of displacement, using the same definition of work given in the section on mechanics. This work can also be expressed in terms of E from the definition of electric field as the ratio of force to charge: W · d, E = F/ q and W = q′. See Figu
Work is change in potential energy: U B − U A = q′ Ed.
In general, the electrostatic potential difference, sometimes called the electric potential difference, is defined as the energy change per unit positive charge, or V B − V A = ( U B − U A )/ q′. For certain configurations of electric field, it may be necessary to use the integral definition of electrostatic potential:
where a test charge moves over a line integral from point A to point B along path s in an electric field ( E).
For the special case of parallel plates:
where V is the potential difference between the plates, measured in units of volts (V):
The electric potential due to a point charge (q) at a distance (r) from the point charge is
The following problem illustrates the calculations of electric field and potential due to point charges.
The first task is to find the region(s) where the electric field is zero. The electric field is a vector, and its direction can be located by a test charge. Figure is divided into three regions. Between the opposite charges, the direction of the force on the test charge will be in the same direction from each charge; therefore, it is impossible to have a zero electric field in Region II. Even though the forces on the test charge from the two charges are in opposite directions in Region I, the force, and therefore the electric field, can never be zero in this region because the test charge is always closer to the largest given charge. Therefore, Region III is the only place where E can be zero. Select an arbitrary point (r) to the right of – Q and set the two electric fields equal. Because the fields are in opposite directions, the vector sum at this point will equal zero.
If X is given, solve for r.
Potential is not a vector, so the potential is zero wherever the following equation holds:
where r m is the distance from the test point to +3 Q and r 2 is the distance to– Q.
This example illustrates the difference in methods of analysis in finding the vector quantity ( E) and the scalar quantity (V). Note that if the charges were either both positive or both negative, it would be possible to find a point with zero electric field between the charges, but the potential would never be zero.
The electrical potential energy of a pair of point charges separated by a distance r is
Equipotential surfaces are surfaces where no work is required to move a charge from one point to another. The equipotential surfaces are always perpendicular to the electric field lines. Equipotential linesare two‐dimensional representations of the intersection of the surface with the plane of the diagram.
Charge
Charge is that property of an object by virtue of which it apply electrostatic force of interaction on other objects.
Charges are of two types
(i) Positive charge
(ii) Negative charge
Like charges repel and unlike charges attract each other.
Quantization of Charge
Charge on any object can be an integer multiple of a smallest charge (e).
Q = ± ne
where, n = 1, 2, 3,……. and e = 1.6 * 10-19 C.
Conservation of Charge
Charge can neither be created nor be destroyed. but can be transferral from one object to another object.
Recently a new particle has been discovered called ‘Quark’. It contains charge ± e / 3, ± 2e / 3.
[The protons and neutrons are combination of other entities called quarks, which have charges 1 / 3 e. However, isolated quarks have not been observed, so, quantum of charge is still e. ]
Coulomb’s Law of Electrostatics
Electrostatic force of interaction acting between two stationary charges is given by
F = 1 / 4π εo q1q2 / r2
where q1, q2 are magnitude of point charges, r is the distance between them and εo is permittivity of free space.
Here, 1 / 4πεo = (10-7 N – s2 / C2)C2
Substituting value of c = 2.99792458 X 108 m/s,
We get 1 / 4πεo = 8.99 x 109N-m2/C2
In examples and problems we will often use the approximate value,
1 / 4πεo = 9 * 109N-m2/C2
The value of εo is 8.85 * 10-12 C2 / N-mC2.
If there is another medium between the point charges except air or vacuum, then εo is replaced by εoK or εoεr or ε.
where K or εr is called dielectric constant or relative permittivity of the medium.
K = εr = ε / εo
where, ε = permittivity of the medium.
For air or vacuum, K = 1
For water, K = 81
For metals, K = ∞
Coulomb’s Law in Vector Form
Force on q2 due to q1,
The above equations give the Coulomb’s law in vector form.
Force on q1 due to q2 = – Force on q2 due to q1
F12 = – F21
F12 = q1q2 / 4πε . r1 – r2 / |r1 – r2|3
The forces due to two point charges are parallel to the line joining point charges; such forces are called central forces and electrostatic forces are conservative forces.
Electric Field
The space in the surrounding of any charge in which its influence can be experienced by other charges is called electric field.
Electric Field Lines
“An electric field line is an imaginary line or curve drawn through a region of space so that its tangent at any point is in the direction of the electric field vector at that point. The relative closeness of the lines at some place give an idea about the intensity of electric field at that point.”
Two lines can never intersect.
Electric field lines always begin on a positive charge and end on a negative charge and do not start or stop in mid space.
Electric Field Intensity (E)
The electrostatic force acting per unit positive charge on a point in electric field is called electric field intensity at that point.
Electric field intensity E =
Its SI unit is NC-1 or Vim and its dimension is [MLT-3 A-1].
It is a vector quantity and its direction is in the direction of electrostatic force acting on positive charge.
Electric field intensity due to a point charge q at a distance r is given by
E = 1 / 4π εo q / r2
Electric Potential (V)
Electric potential at any point is equal to the work done per positive charge in carrying it from infinity to that point in electric field.
Electric potential, V = W / q
Its SI unit is J / C or volt and its dimension is [ML2T-3A-1].
It is a scalar quantity.
Electric potential due to a point charge at a distance r is given by
v = 1 / 4π εo q / r
Potential Gradient
The rate of change of potential with distance in electric field is called potential gradient.
Potential gradient = dV / dr
Its unit is V / m.
Relation between potential gradient and electric field intensity is given by
E = – (dV / dr)
Equipotential Surface
Equipotential surface is an imaginary surface joining the points of same potential in an electric field. So, we can say that the potential difference between any two points on an equipotential surface is zero. The electric lines of force at each point of an equipotential surface are normal to the surface.
(i) Equipotential surface may be planer, solid etc. But equipotential surface can never be point size.
(ii) Electric field is always perpendicular to equipotential surface.
(iii) Equipotential surface due to an isolated point charge is spherical.
(iv) Equipotential surface are planer in an uniform electric field.
(v) Equipotential surface due to a line charge is cylindrical.
Electric Lines of Force
Electric lines of force are the imaginary lines drawn in electric field at Which a positive test charge will move if it is free to do so.
Electric lines of force start from positive charge and terminate on negative charge.
A tangent drawn at any point on electric field represents the direction of electric field at that point.
Two electric lines of force never intersect each other.
Electric lines of force are always perpendicular to an equipotential surface.
Electric Flux (φE)
Electric flux over an area is equal to the total number of electric field lines crossing this area.
Electric flux through a small area element dS is given by
φE = E. dS
where E= electric field intensity and dS = area vector.
Its SI unit is N – m2C-1.
Gauss’s Theorem
The electric flux over any closed surface is 1 / εo times the total charge enclosed by that surface, i.e.,
If a charge q is placed at the centre of a cube, then
total electric flux linked with the whole cube = q / εo
electric flux linked with one face of the cube = q / 6 εo
(i) Electric Field at Any Point on the Axis of a Uniformly Charged Ring A ring-shaped conductor with radius a carries total charge Q uniformly distributed around it. Let us calculate the electric field at a point P that lies on the axis of the ring at distance x from its centre.
Ex = 1 / 4π εo * xQ / (x2 + a2)3/2
The maximum value of electric field
Ex = 1 / 4π εo (2Q / 3√3R2
(ii) Electric Field due to a Charged Spherical Shell
(a) At an extreme point (r > R)
E = 1 / 4π εo q / r2
(b) At the surface of a shell (r = R)
E = 1 / 4π εo q / R2
(c) At an internal point (r < R)
E = 0
(iii) Electric Potential due to a Charged Conducting Spherical Shell
(a) At an extreme point (r > R)
V = 1 / 4π εo q / r
(b) At the surface of a shell (r = R)
V = 1 / 4π εo q / R
(c) At an internal point (r < R)
V = 1 / 4π εo q / R
Therefore potential inside a charged conducting spherical shell equal to the potential at its surface.
(iv) Electric Field and Potential due to a Charged Non-Conducting Sphere
At an extreme point, (r > R)
(v) Electric Field Intensity due to an Infinite Line Charge
E = 1 / 2 π εo λ / r
where λ is linear charge density and r is distance from the line charge.
(vi) Electric Field Near an Infinite Plane Sheet of Charge
E = σ / 2 εo
where σ = surface charge density.
If infinite plane sheet has uniform thickness, then
E = σ / εo
Electric Dipole
An electric dipole consists of two equal and opposite point charges separated by a very small distance. e.g., a molecule of HCL, a molecule of water etc.
Electric Dipole Moment p = q * 2 l
Its SI unit is ‘coulomb-metre’ and its dimension is [LTA).
It is a vector quantity and its direction is from negative charge towards positive charge.
Electric Field Intensity and Potential due to an Electric Dipole
(i) On Axial Line
Electric field intensity E = 1 / 4 π εo * 2 pr / (r2 – a2)2
If r > > 2a, then E = 1 / 4 π εo * 2 p / r3
Electric potential V = 1 / 4 π εo * p / (r2 – a2)
Ifr > > 2a, then V = 1 / 4 π εo * p / r2
(ii) On Equatorial Line
Electric field intensity E = 1 / 4 π εo * p / (r2 + a2)3 / 2
If r > > 2a, then E = 1 / 4 π εo * p / r3
Electric potential V = 0
(iii) At any Point along a Line Making θ Angle with Axis
Electric field intensity E = 1 / 4 π εo * p √1 + 3 cos2 θ / r3
Electric potential V = 1 / 4 π εo * p cos θ / (r2 – a2 cos2 θ)
If r > > 2a, then V = 1 / 4 π εo * p cos θ / r2
Torque
Torque acting on an electric dipole placed in uniform electric field is given by
τ = Ep sin θ or τ = p * E
When θ = 90°, then ‘τmax = Ep
When electric dipole is parallel to electric field, it is in stable equilibrium and when it is anti-parallel to electric field, it is in unstable equilibrium.
Work Done
Work done is rotating an electric dipole in a uniform electric field from angle θ1 to θ2 is given by
W = Ep (cos θ1 – cos θ2)
If initially it is in the direction of electric field, then work done in rotating through an angle θ, W = Ep (1 – cos θ).
Potential Energy
Potential energy of an electric dipole in a uniform electric field is given by U = – pE cos &theta.
Dipole in Non-uniform Electric Field
When an electric dipole is placed in a non-uniform electric field, then a resultant force as well as a torque act on it.
Net force on electric dipole = (qE1 – qE2), along the direction of greater e ecmc field intensity.
Therefore electric dipole undergo rotational as well as linear motion.
Potential Energy of Charge System
Two point charge system, contains charges q1 and q2 separated by a distance r is given by U = 1 / 4 π εo * q1q2 / r
Three point charge system
U = 1 / 4 π εo * [q1q2 / r1 + q2q3 / r2 + q3q2 / r3
Important Points
When charge is given to a soap bubble its size gets increased.
In equilibrium for a charged soap bubble, pressure due to surface tension
= electric pressure due to charging
4T / r = σ2 / 2 εo
or 4T / r = 1 / 2 εo (q / 4 πr2)2
or q = 8 πr √2 εo rT
where, r is radius of soap bubble and T is surface tension of soap bubble.
Behaviour of a Conductor in an Electrostatic Field
(i) Electric field at any point inside the conductor is zero.
(ii) Electric field at any point on the surface of charged conductor is directly proportional to the surface density of charge at that point, but electric potential does not depend upon the surface density of charge.
(iii) Electric potential at any point inside the conductor is constant and equal to potential.
Electrostatic Shielding
The process of protecting certain field from external electric field is called, electrostatic shielding.
Electrostatic shielding is achieved by enclosing that region in a closed metallic chamber.
Dielectric
Dielectrics are of two types Non-polar Dielectric The non-polar dielectrics (like N2, O2, benzene, methane) etc. are made up of non-polar atoms/molecules, in which the centre of positive charge coincides with the centre of negative charge of the atom/molecule.
Polar Dielectric
The polar dielectric (like H2O, CO2, NH3 etc) are made up of polar atoms/molecules, in which the centre of positive charge does not coincide with the centre of negative charge of the atom.
Capacitor
A capacitor is a device which is used to store huge charge over it, without changing its dimensions.
When an earthed conductor is placed near a charged conductor, then it decreases its potential and therefore more charge can be stored over it.
A capacitor is a pair of two conductors of any shape, close to each other and have equal and opposite charges.
Capacitance of a conductor C = q / V
Its 81 unit is coulomb/volt or farad.
Its other units are 1 μ F = 10-6 F
1 μμ F = 1 pF = 10-12 F
Its dimensional formula is [M-1L-2T4A2].
Capacitance of an Isolated Spherical Conductor
C = 4 π εo K R
For air K = 1
∴ C = 4 π εo R = R / 9 * 109
Parallel Plate Capacitor
The parallel plate capacitor consists of two metal plates parallel to each other and separated by a distance d.
Capacitance C = KA εo / d
For air capacitor Co = A εo / d
When a dielectric slab is inserted between the plates partially, then its capacitance.
C = A εo / (d – t + t / K)
If a conducting (metal) slab is inserted between the plates, then
C = A εo / (d – t)
When more than one dielectric slabs are placed fully between the plates, then
The plates of a parallel plate capacitor attract each other with a force
F = Q2 / 2 A εo
When 9. dielectric slab is placed between the plates of a capacitor than charge induced on its side due to polarization of dielectric is
q’ = q (1 – 1 / k)
Capacitors Combination
(i) In Series
Resultant capacitance = 1 / C = 1 / C1 + 1 / C2 + 1 / C3 + ….
In series charge is same on each capacitor, which is equal to the charge supplied by the source.
If V1, V2, V3,…. are potential differences across the plates of the capacitors then total voltage applied by the
source
V = V1 + V2 + V3 + ….
(ii) In Parallel
Resultant capacitance C = C1 + C2 + C3 + ….
In parallel potential differences across the plates of each capacitor is same.
If q1, q2, q3 are charges on the plate of capacitors connected in parallel then total charge given by the source
q = q1 + q2 + q3 + ….
Electric potential energy of a charged conductor or a capacitor is given by,
U = 1 / 2 Vq = 1 / 2 CV2 = 1 / 2 q2 / C
Redistribution of Charge
When two isolated charged conductors are connected to each other then charge is redistributed in the ratio of their capacitances.
Common potential V = q1 + q2 / C1 + C2 = C1V1 + C2V2 / C1 + C2
Energy loss = 1 / 2 C1C2 (V1 – V2)2 / (C1 + C2)
This energy is lost in the form of heat in connecting wires.
When n, small drops, each of capacitance C, charged to potential V with charge q, surface charge density σ and potential energy U coalesce to from a single drop.
Then for new drop
Total charge = nq
Total capacitance = nl/3C
Total potential = n2/3 V
Surface charge density = nl/3 σ
Total potential energy = n2/3 U
Van-de-Graaff Generator
It is a device used to build up very high potential difference of the order of few million volt.
Its working is based on two points
(i) The action of sharp points (corona discharge)
(ii) Total charge given to a spherical shell resides on its outer surface.
Lightning Conductor
When a charged cloud passes by a tall building, the charge on cloud passes to the earth through the building. This causes a damage to the building. Thus to protect the tall building lightning, the lightning conductors, (which are pointed metal roe passes over the charge on the clouds to earth, thus protecting building.
Assignment
ASSIGNMENT : ELECTROSTATICS ASSIGNMENT AND SOLUTIONS MARKS : 10 DURATION : 1 week, 3 days