Answers to Problems on Gauss's Law HC Verma's Questions for Short Answer

Q#1

A small plane area is rotated in an electric field. In which orientation of the area is the flux of the electric field through the area maximum?  

Answer: 

The flux (Δⲫ) of an electric field (E) through an area (Δs) is given as,

Δⲫ = E.Δs = EΔscosß

Where ß is the angle between the direction of the electric field and the direction of normal to the area Δs.

At a fixed place E is constant and the plane area Δs is also fixed. So the flux will be maximum for the maximum cosß. The maximum value of cosß is 1 for ß = 0°.

That is when the direction of E and normal to the area is the same. In other words, the flux is maximum when the plane of the area is perpendicular to the direction of the electric field.       

Q#2

A circular ring of radius r made of a nonconducting material is placed with its axis parallel to a uniform electric field. The ring is rotated about a diameter through 180°. Does the flux of the electric field change? If yes, does it decrease or increase?  

Answer: 

Δs = πr².

In the first case, the axis of the ring is parallel to the direction of the electric field. Since one side of the normal to the area is assumed positive, let us assume that the normal along the electric field is positive.

So, in the first case, the flux of the electric field through the ring is,

 ΔΦ = EΔscos0°

ΔΦ =  E(πr²) = πr²E.

When the ring is rotated through 180°, the angle between the electric field and the normal becomes 180°.

Now the flux is,

 ΔΦ' = EΔscos180° = -πr²E

So, the flux of the electric field changes ant it decreases. 

Q#3

A charge Q is uniformly distributed on a thin spherical shell. What is the field at the center of the shell? If a point charge is brought close to the shell, will the field at the center change? Does your answer depend on whether the shell is conducting or nonconducting? 

Answer: 

Since the charge is uniformly distributed on the thin spherical shell, the field vectors at the center will be symmetrical all around and cancel out. So the net field at the center will be zero.

When a point charge is brought near the shell we have two conditions, 

First when the shell is conducting. A large number of free electrons in the conductor will rearrange themselves and the shell will no longer remain uniformly charged. But the electric field inside the conductor will be zero everywhere even when it is in the field of a point charge and all the charges on the conductor will be on the outer surface. If we take the Gaussian surface through the thickness of the shell, the electric field at all the points of this surface = zero.

So, flux through this surface is zero and there is no charge inside this surface. So, the electric field inside this surface and hence at the center is also zero.

The second condition is when the shell is non-conducting. In this case, there are not large numbers of free electrons to rearrange and there is no surety of the electric field being zero inside the material. Hence the shell will still remain uniformly charged and the field due to it at the center will be zero but the field at the center due to the point charge (q) will be non-zero and equal to kq/r².         

Q#4

A spherical shell made of plastic, contains a charge Q distributed uniformly over its surface. What is the electric field inside the shell? If the shell is hammered to deshape it without altering the charge. will the field inside be changed? What happens if the shell is made of metal?  

Answer: 

Just like in gravitation, the electric field due to a uniformly charged thin spherical shell at an internal point is zero. 

When the non-conducting plastic shell is deformed, the charge distribution on it is disturbed though the total charge is the same. Since it is non-conducting, the field inside its material may not be zero, nor the charge will reside only on the outer surface. In such a case the field inside will not be zero and it will change.    

When the shell is made of metal, it is conducting. There will be large numbers of free electrons and the field inside the metal will be zero. And by taking the Gaussian surface through the metal thickness we can prove that the field inside will not change and will remain zero.

Q#5

A point charge is placed in a cavity in a metal block. If a charge Q is brought outside the metal, will the charge q feel an electric force?  

Answer: 

Due to large numbers of free electrons in the metal block, they rearrange themselves such that the electric field inside the metal is always zero. If we chose a Gaussian surface S such that it is always inside the metal and enclosing cavity, the total charge inside it must be zero.

Since already there is a charge q in the cavity a -q charge must have been induced inside the surface of the cavity. So whatever be the charge outside the metal, there will be zero electric fields in the metal body and the charge q will not be affected.     

Q#6

A rubber balloon is given a charge Q distributed uniformly over its surface. Is the field inside the balloon zero everywhere if the balloon does not have a spherical surface?  

Answer: 

A rubber balloon is non-conducting. When the shape is non-spherical, the symmetry is lost. Also, there is an absence of a large number of free electrons to make the inside of the material of the ballon neutral. So, the field inside the balloon will not be zero everywhere.      

Q#7

It is said that any charge given to a conductor comes to its surface. Should all the protons come to the surface? Should all the electrons come to the surface? Should all the free electrons come to the surface?   

Answer:

As we know an atom has a nucleus made of protons and neutrons and the electrons revolve around this nucleus in different orbits. In a neutral atom, the number of protons and electrons are exactly the same. In a conductor, the atoms or the molecules do not move freely nor the protons in the nucleus. The electrons in the outermost orbit are weakly bonded to the nucleus compared to the inner orbit electrons.

When a negative charge is given to the conductor, a lot of free electrons are transferred to it. Only these free electrons come to the surface. 

When a positive charge is given to the conductor, a lot of weakly bonded electrons are taken away from it. There is a deficiency of electrons in the conductor. This deficiency of electrons appears on the surface of the conductor.