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Exercise Short Questions of Electrostatics, Chapter 12 of 2nd Year Physics

 May 01, 2023     2nd Year Physics, Electrostatics, Exercise Short Questions     No comments   

Q # 1. The potential is constant throughout a given region of space. Is the electric field zero or non zero in this region? Explain.

Ans. The electric field intensity is described by the relation:

\[E\  = \ - \frac{\Delta V}{\Delta r}\]

According to the relation, the electric field is negative gradient of electric potential. If the electric potential is constant throughout given region of space, then change in electric potential \(\Delta V = 0\), hence \(E\  = \ 0\).

Q # 2. Suppose that you follow an electric field line due to a positive point charge. Do electric field and the potential increases or decreases.

Ans.  If we follow an electric field line due to a positive point charge, then it means that we are moving await from point charge. Thus the distance from the charge increases. Due to increase of distance from positive charge, both electric field intensity and electric potential decreases as:

\[E \propto \ \frac{1}{r^{2}}\]

And

\[V \propto \ \frac{1}{r}\]

Q # 3. How can you identify that which plate of capacitor is positively charged?

Ans. The presence of charge on a body is detected by a device called gold leaf electroscope. The leaves of gold leaf electroscope are diverged by giving them negative charge.

  • If the disc is touched with any plate of the charged capacitor and the divergence of the leaves increases, the plate of capacitor is negatively charged
  • If the divergence of leaves decreases, then that plate of capacitor is positively charged.

Q # 4. Describe the force or forces on a positive point charge when placed between parallel plates (i) With similar and equal charges, (ii) With opposite and equal charges

Ans. When a positive point charge is placed between parallel plates with similar and equal charges, then the electric field intensity due to one plate is equal in magnitude but opposite in direction of electric intensity due to other plate. So the value of resultant electric field intensity E is zero. Hence the net force on the positive point charge is zero. Thus it will remain at rest.

When a positive point charge is placed between parallel plates with opposite but equal amount of charge, then electric field intensity due to one plate is equal in magnitude but in same direction of the electric field intensity due to other plate. So the value of resultant electric field intensity is non zero. Hence the point charge will be accelerated towards negative plate.

Q # 5. Electric lines of force never cross. Why?

Ans. Electric lines of force never cross each other. This is because of the reason that electric field intensity has only one direction at any given pint. If the lines cross, electric intensity could have more than one direction which is physically not correct.

Q # 6. If a point charge of mass m is released in a non-uniform electric field with field lines in the same direction pointing, will it make a rectilinear motion.

Ans. A non-uniform field of a positive point charge is shown in the figure:

A point charge q of mass m is placed at any point in the field, it will follow straight or rectilinear path along the field line due to electrostatic repulsive force.

Q # 7. Is E necessarily zero inside a charged rubber balloon if the balloon is spherical. Assume that charge is distributed uniformly over the surface.

Ans. Yes, E is necessarily zero inside a charged rubber balloon if balloon is spherical. If the Gaussian surface is imagined inside charged balloon, then it does not contain any charge i.e., q=0.

Applying Gausses law:

\[\Phi_{e} = \frac{q}{\varepsilon_{0}} = 0\]

Also,

\[\Phi_{e} = E.A\]

Comparing, we have:

\[E.A = 0\]

As \(A \neq 0\), therefore, \(E = 0\)

Hence electric field intensity will be zero inside a spherical balloon.

Q # 8. Is it true that Gauss's law states that the total number of lines of force crossing any closed surface in the outward direction is proportional to the net positive charge enclosed within surface?

Ans. Yes, the above statement is true.

 Electric flux is defined as the measure of number of electric lines of force passing through a certain area. According to Gauss's law, the flux through any close surface is \(\frac{1}{\varepsilon_{0}}\) times the total charged enclosed in it.

Electric flux = \(\frac{1}{\varepsilon_{0}}\) (Total Charge Enclosed)

Electric flux = constant (Total Charge Enclosed)

Electric flux \(\propto\) (Total Charge Enclosed)

Q # 9. Do electrons tends to go to region of high potential or of low potential?

Ans. The electrons being negatively charge particle when released in electric field moves from a region of lower potential (negative end) to a region of high potential (positive end).

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