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Assertion:- Any arbitrary displacement o...

Assertion:- Any arbitrary displacement of charges inside a shell does not introduce any change in the electrostatic field of the outer space.
Reason:- A closed conducting shell divides the entire space into the inner and outer parts which are completely independent of one another in respect of electric fields.

A

If both Assertion `&` Reason are True `&` the Reason is a correct explanation of the Assertion.

B

If both Assertion `&` Reason are True but Reason is not a correct explanation of the Assertion.

C

If Assertion is True but the Reason is False.

D

If both Assertion `&` Reason are False.

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A
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Answer carefully: (a) Two large conducting spheres carrying charges Q_(1) and Q_(2) are brought close to each other. Is the magnitude of electrostatic force between them exactly given by Q_(1),Q_(2)//4pi epsilon_(0)r^(2) , where r is the distance between their centres? (b) If Coulomb’s law involved 1//r^(3) dependence (instead of would Gauss’s law be still true ? (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? (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? (e) We know that electric field is discontinuous across the surface of a charged conductor. Is electric potential also discontinuous there? (f) What meaning would you give to the capacitance of a single conductor? (g) Guess a possible reason why water has a much greater dielectric constant (= 80) than say, mica (= 6).

Assertion:A point charge q is placed near an arbitrary shaped solid conductor as shown in figures.The potential difference between the points A and B within the conductor remain same irrespective of the magnitude of charge q Reason: The electric field inside a solid conductor is zero under electrostatic conditions

When two concentric shells are connected by a thin conducting wire, whole of the charge of inner shell transfers to the outer shell and potential difference between them becomes zero. Surface charge densities of two thin concentric spherical shells are sigma and -sigma respectively. Their radii are R and 2R . Now they are connected by a thin wire. Suppose electric field at a distance r (gt 2R) was E_(1) before connecting the two shells and E_(2) after connecting the two shells, then |E_(2)/E_(1)| is :-

When two concentric shells are connected by a thin conducting wire, whole of the charge of inner shell transfers to the outer shell and potential difference between them becomes zero. Surface charge densities of two thin concentric spherical shells are sigma and -sigma respectively. Their radii are R and 2R . Now they are connected by a thin wire. Suppose electric field at a distance r=(3R)/2 was E_(1) before connecting the two shells and E_(2) after connecting the two shells, then |E_(2)/E_(1)| is :-

A spherical conducting shell of inner radius r_(1) and outer radius r_(2) has a charge Q. ls the electric field inside a cavity (with no charge) zero, even if the shell is not spherical, but has any irregular shape ? Explain.

Assertion:- In a given situation of arrangement of charges, an extra charge is placed outside the Gaussion surface. In the Gauss Theroem ointvecEdvecs=(Q_(in))/(epsilon_(0)).Q_(in) remains unchanged whereas electric field vecE at the site of the element is changed. Reason:- Electric field vecE at any point on the Gaussian surface is due to inside charge only.

A spherical conducting shell of inner radius r_(1) and outer radius r_(2) has a charge Q. (a) A charge q is placed at the centre of the shell. What is the surface charge density on the inner and outer surfaces of the shell? (b) Is the electric field inside a cavity (with no charge) zero, even if the shell is not spherical, but has any irregular shape? Explain.

An electromagnetic wave can be represented by E = A sin (kx- omega t + phi) , where E is electric field associated with wave, According this equation, for any value of x, E remains sinusoidal for -oolt t lt oo . Obviously this corresponds to an idealised situation because radiation from ordinary sources consists of finite size wavetrains. In general, electric field remains sinusoidal only for times of order tau_(c) ' which is called coherence time. In simpler language it means that for times of order tau_(c)' a wave will have a definite phase. The finite value of coherence time could be due to many factors, for example if radiating atom undergoes collision with another atom then wave train undergoes an abrupt phase change or due to the fact that an atom responsible for emitting radiation has a finite life time in the energy level from which it drops to lower energy level, while radiating. Concept of coherence time can be easily understood using young's double slit experiment. Let interference patten is observed around point P at time t , due to superposition of waves emanting from S_(1) and S_(2) at times t =(r_(1))/(c) and (r_(2))/(c) respectively, where r_(1) and r_(2) are the distances S_(1) P & S_(2)P . Obviously if (r_(2)-r_(1))/(c) lt lt tau_(e),{"where" " "c = 3xx10^(8)m//s} then, wavetrain arriving at point P from S_(1) & S_(2) will have a definite phase relationship and an interference pattern of good contranst will be obtained. If coherence time is of order 10^(-10) second and screen is placed at a very large distance from slits in the given figure, then:-

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