Q. 314.3( 4 Votes )

# 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}/4πε_{0}r^{2}, where r is the distance between their centres?

(b) If Coulomb’s law involved 1/r^{3} dependence (instead of 1/r^{2}), 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).

Answer :

(a) According to question, two large conducting spheres with charge Q_{1} and Q_{2} are placed such that distance between their centre is r. Then, the magnitude of electrostatic force between the two spheres is not exactly equal to

Since, the magnitude of electrostatic force between two-point charges Q_{1} and Q_{2} placed at a distance r is given by the relation,

.

Here both the spheres have non-uniform distribution of charges Q_{1} and Q_{2} on their surface.

(b) Gauss’s law states that net electric flux through a surface is the electric field of the charge enclosed by the surface multiplied by the area of the surface projected in a plane perpendicular to the field and this product is a constant equal to charge enclosed within the surface divided by ϵ_{0}

i.e.

or

Where, electric flux,

E = electric field

and dA is the elementary area vector of the surface, Q = charge enclosed by the surface, ϵ_{0} = permittivity of free space.

Since, Electric field involves 1/r^{2} and area vector involves the term of r^{2}, therefore their product becomes constant.

But, If Coulomb’s law involved 1/r^{3} dependence (instead of 1/r^{2}), product of electric field and area vector Gauss’s law will not be true.

(c) If a small test charge is released at rest at a point in an electrostatic field configuration, then it will travel along the field lines passing through that point, as the tangent drawn at a point on the field line gives the direction of acceleration at that point.

(d) When the electron completes an orbit, either circular or elliptical, the displacement becomes zero, so the work done by the field of a nucleus is zero. ∵ Work = force × displacement

(e) Electric field is discontinuous across the surface of a charged conductor, but the electric potential is continuous.

(f) The capacitance of a single conductor means a capacitance of a parallel plate capacitor with one plate as the conductor while the other placed at infinity.

(g) Water has a permanent dipole moment while mica doesn’t, because water has an unsymmetrical space as compared to mica. So, water has a greater dielectric constant than mica.

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Draw the equipotential surfaces due to an isolated point charge.

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