Electrostatic (Previous year Questions)

Q1. If in the electric field, an electron is kept freely. If the electron is replaced by a proton, what will be the relationship between the forces experienced by them?

Ans. When Electron is replaced by proton then the force on proton will remain the same but the direction of the force gets reverse. 

Q2. How does the Coulombian force between two-point charges depend upon the dielectric constant of the intervening medium?

Ans. If medium has some dielectric constant value, then force will be $ \frac{1}{K} $ times the force when medium is air. K is the dielectric constant.

Q3. Draw electrostatic field lines due to a small conducting sphere having a negative charge on it.

Ans.  Electric fields coming out of the charge.

Q4. An electrostatic field line cannot be discontinuous. Why?

Ans. Electric field lines can not be discontinuous because field is continuous in space and exist at all points in space. Click here

Q5. Two electric field lines never cross each other. Why?

Ans. If electric field lines will intersect then there will be two directions of the electric field at a point which is not possible.

Q6. Sketch the pattern of electric field lines due to an electric dipole.

Ans. Click here    

Q7. Which orientation of an electric dipole in a uniform electric field corresponds to stable equilibrium? 

Ans. When the dipole is placed along the direction of the electric field.

Q8. At what points, is the field due to an electric dipole parallel to the line joining the charges?

Ans. At Equitorial point and axial line, Electric field is parallel to dipole.

Q9. If the radius of the Gaussian surface enclosed a surface is halved, how does the electric flux through the Gaussian surface change?

Ans. There will be no change in electric  flux.  

Q10. Electric field inside a conductor is zero, explain.

Ans. Total charge within the conductor is zero so electric field is zero. 

Q11. The electric field E due to a point charge at any point near it is defined as \[E = \lim_{q_0\to 0 } \frac{F}{q_0}\] Where q₀ is the test charge and F is the force acting on it. What is the physical significance of \[\lim_{q_0\to 0 }\] in this expression? 

Ans. The charge is taken as small as possible so that it's presence does not affect the electric field.

Q12. An electric dipole is free two move in a uniform electric field. Explain its motion.

Ans. In a uniform electric field, total force acting on the dipole is zero but torque acting is not zero. Because of the torque,  dipole rotate in the uniform electric field.

Q13. How much work is done in moving a 500 mC charge between two points on an equipotential surface? 

Ans. Work done on moving a charge over equipotential surface is zero irrespective of the magnitude of the charge.

Q14. In which position, a dipole placed in a uniform electric field is in (i)  stable (ii) unstable equilibrium?

Ans. (i) Dipole is stable when placed along the direction of the electric field.

(ii) Dipole is unstable when placed opposite to the direction of the electric field.

Q15. Why does the electric field inside a dielectric decrease when it is placed in an external electric field?

Ans. Electric field decrease because due to external electric field there is induced dipole moment in the dielectric which gives rise to internal electric field that oppose the external field. That is why field decrease inside the dielectric.

Q16. Can two equipotential surfaces intersect each other? Give reasons.

Ans. No, if they intersect then there will be two directions of the electric field which is not possible.

Q17. A parallel plat capacitor is charged to a potential different V by a d.c. source. The capacitor is then disconnected from the source. If a distance between the plate is doubled, state with reason how the following will change; 

(i) electric field between the plates,

(ii) capacitance, and

(iii)energy stored in the capacitor.

Ans. When plate separation is doubled then capacitance will be half of initial value.

(i) Since, Electric field is given by $ E = \frac{\sigma}{\epsilon_0}$ which is independent of the saparation so electric field will be same.

(ii) Since, Capacitance is given by $ C_i = \frac{\epsilon_0 A}{d} $. When d is increased to 2d then C will be half.

(iii) Energy is given by $U = \frac{Q^2}{2C} $. When C will be half then energy will be double of the initial value of energy.