Electrostatics
Branch of science that deals with the study of forces, fields, and potentials arising from the static charges
Electric Charge
- In 600 B.C., the Greek Philosopher Thales observed that amber, when rubbed with wool, acquires the property of attracting objects such as small bits of paper, dry leaves, dust particles, etc.
- This kind of electricity developed on objects, when they are rubbed with each other, is called frictional electricity.
- The American scientist Benjamin Franklin introduced the concept of positive and negative charges in order to distinguish the two kinds of charges developed on different objects when they are rubbed with each other.
- In the table given below, if an object in the first column is rubbed against the object given in second column, then the object in the first column will acquire positive charge while that in second column will acquire negative charge.
- Electric charge − The additional property of protons and electrons, which gives rise to electric force between them, is called electric charge.
- Like charges repel each other whereas unlike charges attract each other.
- A simple apparatus used to detect charge on a body is the gold-leaf electroscope.
Conductors
- The substances which allow electricity to pass through them easily are called conductors.
- Conductors have electrons that can move freely inside the material.
- When some charge is transferred to a conductor, it readily gets distributed over the entire surface of the conductor.
- When a charged body is brought in contact with the earth, all the excess charge on the body disappears by causing a momentary current to pass to the ground through the connecting conductor (such as our body). This process is known as earthing.
- The substances which do not allow electricity to pass through them easily are called insulators.
- Most of the non-metals such as porcelain, wood, nylon, etc. are examples of insulator.
- If some charge is put on an insulator, then it stays at the same place.
A conductor may be charged permanently by induction in the following steps.
Step ITo charge a conductor AB negatively by induction, bring a positively charged glass rod close to it. The end A of the conductor becomes negatively charged while the far end B becomes positively charged. It happens so because when positively charged glass rod is brought near the conductor AB, it attracts the free electrons present in the conductor towards it. As a result, the electron accumulates at the near end A and therefore, this end becomes negatively charged and end B becomes deficient of electrons and acquires positive charge.Step IIThe conductor is now connected to the earth. The positive charges induced will disappear. The negative induced charge on end A of the conductor remains bound to it due to the attractive forces exerted by the positive glass rod.Step IIIThe conductor is disconnected from the earth keeping the glass rod still in its position. End A of the conductor continues to hold the negative induced charge.Step IVFinally, when the glass rod is removed, the negative induced charge on the near end spreads uniformly over the whole conductor.
- Additive nature of charges − The total electric charge on an object is equal to the algebraic sum of all the electric charges distributed on the different parts of the object. Ifq1, q2, q3, … are electric charges present on different parts of an object, then total electric charge on the object, q = q1 + q2 + q3 + …
- Charge is conserved − When an isolated system consists of many charged bodies within it, due to interaction among these bodies, charges may get redistributed. However, it is found that the total charge of the isolated system is always conserved.
- Quantization of charge − All observable charges are always some integral multiple of elementary charge, e (= ± 1.6 × 10−19 C). This is known as quantization of charge.
- Two point charges attract or repel each other with a force which is directly proportional to the product of the magnitudes of the charges and inversely proportional to the square of the distance between them.
Where, [In SI, when the two charges are located in vacuum]
− Absolute permittivity of free space = 8.854 × 10−12 C2 N−1 m−2
We can write equation (i) as
- The force between two charges q1 and q2 located at a distance r in a medium may be expressed as
The ratio is denoted by εr, which is called relative permittivity of the medium with respect to vacuum. It is also denoted by k, called dielectric constant of the medium.
ε = kε0
Coulomb’s Law in Vector Form
Consider two like charges q1 and q2 present at points A and B in vacuum at a distance r apart.
According to Coulomb’s law, the magnitude of force on charge q1 due to q2 (or on charge q2 due to q1) is given by,
Let
− Unit vector pointing from charge q1 to q2
− Unit vector pointing from charge q2 to q1
[is along the direction of unit vector ] …(ii)
[is along the direction of unit vector] …(iii)
∴Equation (ii) becomes
On comparing equation (iii) with equation (iv), we obtain
Forces between Multiple Charges
Principle of superposition − Force on any charge due to a number of other charges is the vector sum of all the forces on that charge due to the other charges, taken one at a time. The individual forces are unaffected due to the presence of other charges.
Consider that n point charges q1, q2, q3, … qn are distributed in space in a discrete manner. The charges are interacting with each other. Let the charges q2, q3, … qn exert forceson charge q1. Then, according to principle of superposition, the total force on charge q1 is given by,
If the distance between the charges q1 and q2 is denoted as r12; and is unit vector from charge q2 to q1, then
Similarly, the force on charge q1 due to other charges is given by,
Substituting these in equation (i),
Electric Field − It is the space around a charge, in which any other charge experiences an electrostatic force.Electric Field Intensity − The electric field intensity at a point due to a source charge is defined as the force experienced per unit positive test charge placed at that point without disturbing the source charge.Where,→ Electric field intensityForce experienced by the test charge q0Its SI unit is NC−1.Electric Field Due To a Point ChargeWe have to find electric field at point P due to point charge +q placed at the origin such thatTo find the same, place a vanishingly small positive test charge q0 at point P.According to Coulomb’s law, force on the test charge q0 due to charge q isIf is the electric field at point P, thenThe magnitude of the electric field at point P is given by,Representation of Electric FieldElectric Field Due To a System of ChargesConsider that n point charges q1, q2, q3, … qn exert forces on the test charge placed at origin O.Let be force due to ith charge qi on q0. Then,Where, ri is the distance of the test charge q0 from qiThe electric field at the observation point P is given by,If is the electric field at point P due to the system of charges, then by principal of superposition of electric fields,Using equation (i), we obtainElectric Field LinesAn electric line of force is the path along which a unit positive charge would move, if it is free to do so.- Properties of Electric Lines of Force
- These start from the positive charge and end at the negative charge.
- They always originate or terminate at right angles to the surface of the charge.
- They can never intersect each other because it will mean that at that particular point, electric field has two directions. It is not possible.
- They do not pass through a conductor.
- They contract longitudinally.
- They exert a lateral pressure on each other.
- Representation of Electric Field Lines
- Field lines in case of isolated point charges
- Field lines in case of a system of two charges
Continuous Charge Distribution
- Linear charge density − When charge is distributed along a line, the charge distribution is called linear.
Where,λ → Linear charge densityq → Charge distributed along a lineL → Length of the rod- Surface charge density
Where,σ → Surface charge densityq → Charge distributed on area A- Volume charge density
Where,δ → Volume charge densityVElectric Dipole − System of two equal and opposite charges separated by a certain small distance.
Electric Dipole Moment − It is a vector quantity, with magnitude equal to the product of either of the charges and the length of the electric dipoleIts direction is from the negative charge to the positive charge.Electric Field on Axial Line of an Electric DipoleLet P be at distance r from the centre of the dipole on the side of charge q. Then,Where, is the unit vector along the dipole axis (from − q to q). Also,The total field at P isFor r >> aElectric Field for Points on the Equatorial PlaneThe magnitudes of the electric field due to the two charges +q and −q are given by,The directions of E+q and E−q are as shown in the figure. The components normal to the dipole axis cancel away. The components along the dipole axis add up.∴ Total electric field[Negative sign shows that field is opposite to]At large distances (r >> a), this reduces toDipole in a Uniform External FieldConsider an electric dipole consisting of charges −q and +q and of length 2a placed in a uniform electric field making an angle θ with electric field.Force on charge −q at (opposite to)Force on charge +q at (along)Electric dipole is under the action of two equal and unlike parallel forces, which give rise to a torque on the dipole.τ = Force × Perpendicular distance between the two forcesτ = qE (AN) = qE (2a sin θ)τ = q(2a) E sinθτ = pE sinθ
Electric Flux Gauss LawThe electric flux, through a surface, held inside an electric field represents the total number of electric lines of force crossing the surface in a direction normal to the surface.Electric flux is a scalar quantity and is denoted by Φ.SI unit − Nm2 C−1Gauss TheoremIt states that the total electric flux through a closed surface enclosing a charge is equal to times the magnitude of the charge enclosed.However,∴Gauss theorem may be expressed asProofConsider that a point electric charge q is situated at the centre of a sphere of radius ‘a’.According to Coulomb’s law,Where, is unit vector along the line OPThe electric flux through area element is given by,Therefore, electric flux through the closed surface of the sphere,It proves the Gauss theorem in electrostatics.
q → Charge on conductorApplications of Gauss Law- Electric Field Due To A Line Charge
Consider a thin infinitely long straight line charge of linear charge density λ.Let P be the point at a distance ‘a’ from the line. To find electric field at point P, draw a cylindrical surface of radius ‘a’ and length l.If E is the magnitude of electric field at point P, then electric flux through the Gaussian surface is given by,- Φ = E × Area of the curved surface of a cylinder of radius r and length lBecause electric lines of force are parallel to end faces (circular caps) of the cylinder, there is no component of field along the normal to the end faces.
Φ = E × 2πal … (i)According to Gauss theorem, we haveFrom equations (i) and (ii), we obtain- Electric Field Due To An Infinite Plane Sheet Of Charge
Consider an infinite thin plane sheet of positive charge having a uniform surface charge density σon both sides of the sheet. Let P be the point at a distance ‘a’ from the sheet at which electric field is required. Draw a Gaussian cylinder of area of cross-section A through point P.The electric flux crossing through the Gaussian surface is given by,- Φ = E × Area of the circular caps of the cylinderSince electric lines of force are parallel to the curved surface of the cylinder, the flux due to electric field of the plane sheet of charge passes only through the two circular caps of the cylinder.
Φ = E × 2A … (i)According to Gauss theorem, we haveHere, the charge enclosed by the Gaussian surface,q = σAFrom equations (i) and (ii), we obtain- Electric Field Due To A Uniformly Charged Thin Spherical Shell
- When point P lies outside the spherical shell
Suppose that we have to calculate electric field at the point P at a distance r (r > R) from its centre. Draw the Gaussian surface through point P so as to enclose the charged spherical shell. The Gaussian surface is a spherical shell of radius r and centre O.Let be the electric field at point P. Then, the electric flux through area element is given by,Since is also along normal to the surface,dΦ = E ds∴ Total electric flux through the Gaussian surface is given by,Now,Since the charge enclosed by the Gaussian surface is q, according to Gauss theorem,From equations (i) and (ii), we obtain- When point P lies inside the spherical shell
In such a case, the Gaussian surface encloses no charge.According to Gauss law,E × 4πr2 = 0i.e., = E = 0 (r < R)
0 comments:
Post a Comment