Two Marks Questions with Answers

Two Marks Questions with Answers

 

Q.1 Define current density and state its unit.

Ans. : The current density is defined as the current passing through the unit surface area, when the surface is held normal to the direction of the current. The current density is measured in amperes per square metres (A/m²).

 

Q.2 State the relation between current and current density. Also state how current density is relate to charge density.

Ans. :

The current I and current density   are related as,

where dS is differential surface area.

The relation between  and ρv can be expressed as,

 

Q.3 State the continuity equation in integral and differential form.

AU: Dec.-08, May-12

Ans. : Let Q_i = Charge within the closed surface

dQ_i / dt = Rate of decrease of charge inside the closed surface

The negative sign indicates decrease in charge.

Due to principle of conservation of charge, this rate of decrease is same as rate of outward flow of charge, which is a current.

This is the integral form of the continuity equation of the current.

The current or the charge per second, diverging from a small volume per unit volume is equal to the time rate of decrease of charge per unit volume at every point.

Hence,

This is the point form or differential form of the continuity equation of the current

 

Q.4 What is mobility ? Obtain its units.

Ans. : The drift velocity is directly proportional to the applied electric field. The constant of proportionality is called mobility of the electrons in a given material and denoted as µe. It is positive for the electrons.

Thus mobility is measured in square metres per volt-second (m²/V-s).

 

Q.5 Define Ohm's law at a point.

AU ; Dec.-02, May-04, 05

Ans. : The point form of Ohm's law is given by, For a metallic conductor,

 where

σ = Conductivity of the material

 

Q.6 State the effect of temperature on resistivity and conductivity.

Ans. : The resistivity is the reciprocal of the conductivity. The conductivity depends on the temperature. As the temperature increases, the vibrations of crystalline structure of atoms increases. Due to increased vibrations of electrons, drift velocity decreases, hence the mobility and conductivity decreases. So as temperature increases, the conductivity decreases and resistivity increases.

 

Q.7 State the Ohm's law for nonuniform fields.

Ans. : For nonuniform fields, the resistance R is defined as the ratio V to I where V is the potential difference between two specified equipotential surfaces in the material and I is the current crossing the more positive surface of the two, into the material. Mathematically the resistance for nonuniform fields is given by,

The numerator is a line integration giving potential difference across two ends while the denominator is a surface integration giving current flowing through the material.

 

Q.8 State the properties of conductor.

(Refer section 5.4.3.)     

 

Q.9 What do you understand from continuity equation ?

Ans.  : If the current flows outwards from the closed surface, then it is constituted due to outward flow of positive charges from the closed surface S. The continuity equation tell that, there must be decrease of an equal amount of positive charge inside the closed surface. Hence the outward rate of flow of positive charge gets balanced by the rate of decrease of charge inside the closed surface. While the differential form of continuity equation states that the current or the charge per second, diverging from a small volume per unit volume is equal to the time rate of decrease of charge per unit volume at every point.

 

Q.10 What is the difference between homogeneous and nonhomogeneous medium ?

Ans. : The medium is called homogeneous when the physical characteristics of the medium do not vary from point to point but remain same everywhere throughout the medium. If the characteristics vary from point to point, the medium is called nonhomogeneous or heterogeneous.

 

Q.11 What is the difference between linear and nonlinear medium ?

Ans. : The medium is called linear with respect to the electric field if the flux density   is directly  proportional to the electric field . The relationship is through the permittivity of the medium. If   is not directly proportional to , the material is called nonlinear.

 

Q.12 What is polarization ?

Ans. : The atoms of a dielectric are electrically neutral. The nucleus is positively charged while negative charge is considered to be in the form of cloud of electrons around the nucleus. Without application of electric field, the charge distribution is symmetric. When electric field  is applied, the symmetrical distribution of charges gets disturbed. The positive charges experience a force  while the negative charges experience a force  in the opposite direction. Now there is separation between the nucleus and the centre of the electron cloud. This forms an electric dipole. The dipole gets aligned with the applied field. This process is called polarization of dielectrics.

 

Q.13 State the mathematical equation for polarization and state its unit.

Ans. : The polarization is defined as the total dipole moment per unit volume.

It is measured in coulombs per square metre (C/m² )

 

Q.14 State the properties of dielectric materials. (Refer section 5.6.3)

AU : May-07

 

Q.15 Define dielectric strength. Give its units. State its value for the air. AU: Dec.-06, 17, May-10

Ans.: The minimum value of the applied electric field at which the dielectric breaks down is called dielectric strength of that dielectric.

The dielectric strength is measured in V/m or kV/cm. It also can be stated as the maximum value of electric field under which a dielectric can sustain without breakdown.

The dielectric strength of air is 3 kV/mm.

 

Q.16 Why is the electrostatic potential continuous at a boundary ?

AU: Dec.-03

Ans. : At the boundary,

E _tan1 = E _{tan 2}

Thus the tangential components of the field intensity at the boundary in both the dielectrics remain same. Thus electric field intensity is continuous at the boundary. And  hence the electrostatic potential is also continuous at the boundary.

 

Q.17 Calculate the energy stored in a 10 µF capacitor which has been charged to a voltage of 400 V.

AU : May-04

Ans.: E = 1/2 CV² =-2 × 10 × 10^-6 × (400) ² = 0.8 J

 

Q.18 Define the boundary conditions for the conductor - free space boundary in electrostatic.

AU : May-03, 04, Dec.-05

Ans. : The boundary conditions for the conductor - free space boundary in electrostatic are,

E_tan = 0 and D_tan = Ɛ₀ E_tan = 0

D_N = ρS and E_N = ρS| / Ɛ₀

 

Q.19 Using Gauss’s law, derive the capacitance of a co-axial cable.

Ans. : According to Gauss's law, for a co-axial cable

 

Q.20 Define the boundary conditions at the interface between two dielectrics.

AU: May-05,18

Ans.

The boundary conditions for the boundary between two perfect dielectrics are,

 

 

Q.21 Explain the electric field distribution inside and outside a conductor.

Ans. : It is known that current must flow when potential gradient exists. When a certain amount of electric charge exists inside a conductor, then mutual repulsion between the charges takes place and it gives rise to an electric field  . This produces the current according to Ohm's law. But as the charges get distributed uniformly,   vanishes and the current ceases. Thus   is everywhere zero inside a conductor and all points inside the conductor are equipotential in nature.

When motion of the charges inside the conductor stops, is zero everywhere inside the conductor. If a Gaussian surface is selected just inside the surface of the conductor as shown in the Fig. 5.18.1 no charge gets enclosed as  is zero.

But according to divergence theorem the volume integral  dv must be zero.

 Thus in the entire volume occupied by the conductor, no charge exists. If somehow charge is released inside the conductor, it immediately flows out and reside on the surface of the conductor. Hence electric charge can reside only on the surface of the conductor.

The  at all points on the surface of a conductor is always normal to the surface. This is because charges on the surface are at rest. If  is not normal but inclined, it will produce Etan component. This can produce the current along the surface i.e. movement of charges. But practically there cannot be any movement of charges on the surface. Hence  must be normal to the surface.

 

Q.22 How is the principle of conservation of charges depicted ?

Ans. : The principle of conservation of charge states that charges can neither be created nor be destroyed though equal amounts of positive and negative charges may be simultaneously created, obtained by method of separation or by destruction or lost by recombination.

 

Q.23 What is capacitor ? What is capicitance ?

AU : May-16

Ans. : A system which has two conducting surfaces carrying equal and opposite charges, separated by a dielectric is called capacitor giving rise to a capacitance.

The ratio of the magnitudes of the total charge on any one of the two conductors and potential difference between the conductors is called the capacitance of the two conductor system denoted as C. The capacitance is the property of capacitor to store electrical energy in the form of electrostatic field when potential difference is applied across it.

 

Q.24 Define 1 farad.

Ans. : One farad of capacitance is defined as the capacitance of a capacitor which requires a charge of one coulomb to establish the potential difference of one volt between its plates.

 

Q.25 State the capacitance of a co-axial cable. Ans. : The capacitance of a co-axial cable is,

where L is its length, a = inner radius and b = outer radius.

 

Q.26 State the capacitance of a spherical capacitor.

Ans. : The capacitance of a spherical capacitor is,.

where the radius of outer sphere is 'b' while that of inner sphere is 'a'.

 

Q.27 State the expression for energy stored and energy density in a capacitor.

Ans. : The energy stored in a capacitor is given by,

While the energy density is given by,

 

Q.28 Determine the capacitance of a parallel plate capacitor having tin foil sheets, 25 cm square plates separated through a glass dielectric 0.5 cm thick with relative permitivity 6.

Ans. :

 

Q.29 Derive point form of Ohms law from V = IR.

Ans. : The Ohm's law is V = IR

AU : Dec.-10

 

Q.30 A parallel plate capacitor has voltage of 25 V across the plates. If the distance between the plates is doubled, find the new voltage across the plates.

Ans. :

 

Q.31 Calculate the current density of copper wire having conductivity of 5.8 × 10⁷ S/m and |  | =20 V/m.

Ans.:

 

Q.32 Determine the value of capacitance between 2 square plates cross sectional area 1 sq.cm separated by 1 cm placed in a liquid whose dielectric constant is 6 and the relative permittivity of free space is 8.854 picoF/m.

AU : May-04

Ans. :

 

Q.33 At the boundary between copper and aluminium the electric field lines make an angle of 45° with the normal to the interface. Find the angle of emergence. The conductivity of copper and aluminium are 5.8 x105 S/cm and 3.5× 10⁵ S/cm, respectively.

Ans. : For the two conductors, boundary condition states that,

 

Q.34 A parallel plate capacitor has a charge of 10-3 C on each plate while the potential difference between the plates is 1000 V. Calculate the value of capacitance.

Ans. :

 

Q.35 A dielectric slab of flat surface with relative permittivity 4 is disposed with its surface nromal to a uniform field with flux density 1.5 C/m2. The slab is uniformly polarized. Determine polarization in the slab.

Ans. :

 

Q.36 If a dielectric material of Ɛ_R = 4.0 is kept in an electric field , find the polarization.

Ans. :

 

Q.37 Calculate the capacitance of a parallel plate capacitor having an electrode area of 100 cm2. The distance between the electrodes is 4 mm and the dielectric used has a permittivity of 3.5. The applied potential is 100 Volts.

AU : May-14

Ans. :  A = 100 cm², d = 4 mm, E_r = 3.5

 

Q.38 Calculate the capacitance per km between a pair of parallel wires each of diameter 1 cm at a spacing of 50 cms.

AU : Dec.-15

 

Q.39 Find the capacitance of an isolated spherical shell of radius a.

Ans. : Capacitance of an isolated sphere can be obtained by using b = ∞ in the capacitance of a spherical capacitor.

 

Q.40 Evaluate the capacitance of a single isolated sphere of 1.5 m diameter in free space.

AU : Dec.-18

Ans. :

 

Q.41 Why water has much greater dielectric constant than mica ?

AU : May-19

Ans . :

The water has a polar nature and has a very large electric dipole moment which is parmanent in nature. The mica has induced dipole moment hence the water has much greater dielectric constant than mica.

 

Q.42 How is electric energy stored in a capacitor ?

Ans. : In a capacitor, the work done in charging a capacitor is stored in the form of an electrostatic energy.