Electric charges and fields

 

                           Electric Charges And Fields


                                          Problems

Question 1. Draw a diagram to show lines of force in a plane containing two equal point charges of opposite sign separated by a small distance. Giving reason, indicate on the diagram a point where a small positive charge experiences a force parallel to the line joining the two charges.



Solution. As shown in the above figure, electric field at any point P on the equatorial line is parallel to the line joining the two charges. So if a small positive charge is placed at such a point, it will experience a force parallel to the line join in the two charges.

Question 2. What is meant by the statement that the electric field of a point charge has spherical symmetry whereas that of an electric dipole is cylindrical symmetric?

Solution. The electric field due to a point charge q at distance r from it is given by

                

Question 3. An electric dipole free to move is placed in a uniform electric field. Explain along with diagram its motion when it is placed,

(a) Parallel to the field.

(b) Perpendicular to the field.

Solution. (a) Since the line of action of the two forces passes through the same point, the net force and the net torque acting on the dipole is zero. So no motion is produced when a dipole is placed to the electric field.









Solution: The directions of the two dipole moments and their resultant are shown in the above figure.

 

Question 7. (a) Define electric flux. Write its SI Units. A spherical balloon carries a charge that is uniformly distributed over its surface. As the balloon us blown up and increases in size, how does the total electric flux coming out of the surface change? Give reason.

Solution: (a) The electric flux through a given surface area is the total number of electric lines of force passing normally through that area.

It is given by.

 As the balloon is blown up and increases in size, the total charge on its surface remains constant. Hence, by Gauss’s theorem, the electric flux coming out of its surface remains unchanged.






Find

(i)The force on the charge at the centre of shell and at the point A,

(ii)The electric flux through the shell.

Solution. Net force on the charge Q/2 placed at the centre of the shell is zero.

Force in the charge 2Q kept at point A at distance r from the centre is


Question 9. Two thin concentric and co-planar spherical shells, of radii a and b (b ˃ a) carry charges q and Q, respectively. Find the magnitude of the electric field, at a point distant from their common centre for 

(i) 0 <  < a         (ii) a < b   (iii) b  <


Solution. (i) For 0 <  < a, the charge enclosed by Gaussian surface I is zero.




By Gauss’s theorem,



(i) For a < < b, the net charge enclosed by the Gaussian II is q.


(ii)For b  <  the net charge enclosed by the Gaussian surface III is (q + Q).


Question 10. A spherical rubber balloon carries a charge that is uniformly distributed over its surface. As the balloon is blown up ; how does E vary for points (i) inside the balloon, (ii) on the surface of the balloon and (iii) outside the balloon ?

Solution. (i) For points inside the balloon, E=0.

(iii) For points outside the balloon,

As the balloon is blown up, the charge enclosed by the Gaussian surface remains same, so E does not change.


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