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2.36 Answer the following:

What are the forms of energy into which the electrical energy of the atmosphere is dissipated during a lightning? (Hint: The earth has an electric field of about 100 Vm^{-1}  at its surface in the downward direction, corresponding to a surface charge density = -10^{-9}Cm^{-2}. Due to the slight conductivity of the atmosphere up to about 50 km (beyond which it is good conductor), about + 1800 C is pumped every second into the earth as a whole. The earth, however, does not get discharged since thunderstorms and lightning occurring continually all over the globe pump an equal amount of negative charge on the earth.) 

Electrical energy, of the atmosphere, is dissipated as light energy which comes from lightning, heat energy and sound energy which comes from the thunderstorm. 

2.36 Answer the following:

c) The discharging current in the atmosphere due to the small conductivity of air is known to be 1800 A on an average over the globe. Why then does the atmosphere not discharge itself completely in due course and become electrically neutral? In other words, what keeps the atmosphere charged? 

Thunderstorm and lightning across the globe keep the atmosphere charged by releasing the light energy, heat energy, and sound energy in the atmosphere. In a way or other, the atmosphere is discharged through regions of ordinary weather. on an average, the two opposing currents are in equilibrium. Hence the atmosphere perpetually remains charged. 

2.36 Answer the following:

(b) A man fixes outside his house one evening a two metre high insulating slab carrying on its top a large aluminium sheet of area 1m2. Will he get an electric shock if he touches the metal sheet next morning?

Yes, the man will get an electric shock. the aluminium sheet is gradually charged up by discharging current of atmosphere.eventually the voltage will increase up to a certain point depending on the capacitance of the capacitor formed by aluminium sheet, insulating slab and the ground. When the man touches the that charged metal, he will get a shock. 

2.36 Answer the following:

(a) The top of the atmosphere is at about 400 kV with respect to the surface of the earth, corresponding to an electric field that decreases with altitude. Near the surface of the earth, the field is about 100Vm^{-1}. Why then do we not get an electric shock as we step out of our house into the open? (Assume the house to be a steel cage so there is no field inside!)

The surface of the earth and our body, both are good conductors. So our body and the ground both have the same equipotential surface as we are connected from the ground. When we move outside the house, the equipotential surfaces in the air changes so that our body and ground is kept at the same potential. Therefore we do not get an electric shock.

2.35 A small sphere of radius rand charge q is enclosed by a spherical shell of radius r2 and charge q2. Show that if q1 is positive, charge will necessarily flow from the sphere to the shell (when the two are connected by a wire) no matter what the charge q2 on the shell is.

The potential difference between the inner sphere and shell; So,  the potential difference  is independent of . And hence whenever q1 is positive, the charge will flow from sphere to the shell

2.34 Describe schematically the equipotential surfaces corresponding to

(d) a uniform grid consisting of long equally spaced parallel charged wires in a plane.

 

The equipotential surface near the grid is periodically varying.and after long distance it becomes parallel to the grid.

2.34 Describe schematically the equipotential surfaces corresponding to

(c) a single positive charge at the origin, and

For a single positive charge, the equipotential surface will be the sphere with centre at position of the charge which is origin in this case.

2.34  Describe schematically the equipotential surfaces corresponding to

(b) a field that uniformly increases in magnitude but remains in a constant (say, z) direction

The potential in a direction perpendicular to the direction of the field is always gonna be same irrespective of the magnitude of the electric field. Hence equipotential surface will be the plane, normal of which is the direction of the field.

2.34 Describe schematically the equipotential surfaces corresponding to

(a) a constant electric field in the z-direction

When the electric field is in the z-direction is constant, the potential in a direction perpendicular to z-axis remains constant. In other words, every plane parallel to the x-y plane is an equipotential plane. 

2.33 A parallel plate capacitor is to be designed with a voltage rating 1 kV, using a material of dielectric constant 3 and dielectric strength about 10^{-7}Vm^{-1}. (Dielectric strength is the maximum electric field a material can tolerate without breakdown, i.e., without starting to conduct electricity through partial ionisation.) For safety, we should like the field never to exceed, say 10% of the dielectric strength. What minimum area of the plates is required to have a capacitance of 50 pF?

Given Voltage rating in designing capacitor The dielectric constant of the material  Dielectric strength of material =  Safety Condition: The capacitance of the plate  Now, As we know, Now,  Hence the minimum required area is 

2.32 A cylindrical capacitor has two co-axial cylinders of length 15 cm and radii 1.5 cm and 1.4 cm. The outer cylinder is earthed and the inner cylinder is given a charge of 3.5\muC. Determine the capacitance of the system and the potential of the inner cylinder. Neglect end effects (i.e., bending of field lines at the ends).

Given Length of cylinder  inner radius  outer radius  Charge on the inner cylinder  Now as we know, The capacitance of this system is given by  Now Since the outer cylinder is earthed the potential at the inner cylinder is equal to the potential difference between two cylinders. SO Potential of inner cylinder:

2.31 Answer carefully:

(g) Guess a possible reason why water has a much greater dielectric constant (= 80) than say, mica (= 6).

Water has a much greater dielectric constant than mica because it posses a permanent dipole moment and has an unsymmetrical shape.

2.31 Answer carefully:

(f) What meaning would you give to the capacitance of a single conductor?

There is no meaning in the capacitor with a single plate factually. but we give it meaning by assuming the second plate at infinity. Hence capacitance of a single conductor is the amount of change required to raise the potential of the conductor by one unit amount.

2.31 Answer carefully:

(e) We know that electric field is discontinuous across the surface of a charged conductor. Is electric potential also discontinuous there?

Since the electric potential is not a vector quantity unlike the electric field, it can never be discontinuous.

2.31 Answer carefully:

c) A small test charge is released at rest at a point in an electrostatic field configuration. Will it travel along the field line passing through that point?

when a small test charge is released at rest at a point in an electrostatic field configuration  it travels along the field line passing through that point only if the field lines are straight because electric field lines give the direction of acceleration, not the velocity

2.31 Answer carefully:

 (d) What is the work done by the field of a nucleus in a complete circular orbit of the electron? What if the orbit is elliptical?

The initial and final position will be the same for any orbit whether it is circular or elliptical. Hence work done will always be zero. 

2.31 Answer carefully: 

(b) If Coulomb’s law involved \frac{1}{r^{3}} dependence (instead of \frac{1}{r^{2}} ), would Gauss’s law be still true ?

Since the solid angle is proportional to  and not proportional to , The guess law which is equivalent of coulombs law will not hold true.   

2.31 Answer carefully:

(a) Two large conducting spheres carrying charges Q1 and Q2 are brought close to each other. Is the magnitude of electrostatic force between them exactly given by \frac{Q_{1}Q_{2}}{4\pi \epsilon_{0}r^{2}} , where r is the distance between their centres?

 

The charge on the sphere is not exactly a point charge, we assume it when the distance between two bodies is large. when the two charged sphere is brought closer, the charge distribution on them will no longer remain uniform. Hence it is not true that electrostatic force between them exactly given by  . 

2.30 A spherical capacitor has an inner sphere of radius 12 cm and an outer sphere of radius 13 cm. The outer sphere is earthed and the inner sphere is given a charge of 2.5 \muC. The space between the concentric spheres is filled with a liquid of dielectric constant 32.

(c) Compare the capacitance of this capacitor with that of an isolated sphere of radius 12 cm. Explain why the latter is much smaller.

The radius of the isolated sphere  Now, Capacitance of sphere: ON comparing it with the concentric sphere, it is evident that it has lesser capacitance.this is due to the fact that the concentric sphere is connected to the earth. Hence the potential difference is less and capacitance is more than the isolated sphere.  

2.30 A spherical capacitor has an inner sphere of radius 12 cm and an outer sphere of radius 13 cm. The outer sphere is earthed and the inner sphere is given a charge of 2.5 \mu C. The space between the concentric spheres is filled with a liquid of dielectric constant 32. 

(b) what is the potential of the inner sphere?

Potential of the inner sphere is given by  Hence the potential of the inner sphere is .
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