Questions OBJECTIVE - II and Answer Gauss's Law HC Verma Part II

Q#1
Mark the correct options:

(a) Gauss law is valid only for symmetrical charge distributions.

(b) Gauss law is valid only for charges placed in the vacuum.

(c) The electric field calculated by Gauss's law is the field due to the charges inside the Gaussian surface.

(d) The flux of the electric field through a closed surface due to all the charges is equal to the flux due to the charges enclosed by the surface.            

Answer: (d) 

Gauss Law is valid for symmetrical as well as non-symmetrical charge distributions. There is no condition for Gauss law to be true only in a vacuum. Hence the options (a) and (b) are wrong. 

Gauss law refers to the net electric field which is the electric field due to inside charges as well as outside charges of a Gaussian surface. So the option (c) is also not true.

Since the solid angle subtended by a closed surface at an outer point is zero, the flux of all outer charges through the closed Gaussian surface is zero. Hence the flux of electric field through a closed surface due to all charges is equal to the flux due to the charges enclosed by the surface. Option (d) is correct. 

Q#2   

A positive point charge Q is brought near an isolated metal cube.

(a) The cube becomes negatively charged.

(b) The cube becomes positively charged.

(c) The interior becomes positively charged and the surface becomes negatively charged.

(d) The interior remains charge free and the surface gets nonuniform charge distribution.                 

Answer: (d) 

There is no change in the total charge of the metal cube due to the positive charge. Since there are numerous loose electrons in a metal piece, these loose electrons rearrange themselves to make the electric field inside the metal due to the outer charge to zero. Hence the options (a), (b), and (c) are not true. 

Due to the rearrangement of loose electrons, the charges come to the surface of the metal cube and polarize. The interior becomes charge-free but the surface gets nonuniform charge distribution. Option (d) is correct.        

Q#3

A large nonconducting sheet M is given a uniform charge density. Two uncharged small metal rods A and B are placed near the sheet as shown in figure(30-Q4). 

(a) M attracts A

(b) M attracts B

(c) A attracts B

(d) B attracts A.         

Answer: All 

The direction of the electric field due to the sheet near the blocks will be horizontal. Its direction will depend on the nature of the charge on the sheet. Suppose the charge on the sheet is positive, then the direction of the field will be from the sheet towards blocks as shown in the figure. 

The Figure for Q#3

The loose electrons on the metal blocks will rearrange themselves to make the electric field inside equal and opposite to the field due to the sheet. For this, the ends of the blocks A and B facing the sheet will be negatively charged while the ends away from the sheet will get positively charged (please see the figure). Since the negatively charged ends of the blocks are nearer to the positively charged sheet, M attracts both A and B. Between the blocks, nearer ends have the opposite charges. Hence both will attract each other. Thus all options are true.   

Q#4

If the flux of the electric field through a closed surface is zero, 

(a) the electric field must be zero everywhere on the surface

(b) the electric field may be zero everywhere on the surface

(c) the charge inside the surface must be zero

(d) the charge in the vicinity of the surface must be zero.           

Answer: (b), (c). 

From the Gauss law, the flux through a closed surface,

$\phi =\frac{q}{\epsilon_0}$, where q is the net inside charge.

Given here that $\phi$ = 0, hence q = 0.

So the net inside charge must be zero. Option (c) is true. 

Here net charge inside must be zero but there may be different positive and negative charges distributed unevenly sum of which is zero. So the electric field may not be zero everywhere on the surface and there may be a charge inside in the vicinity of the surface. The option (a) and (d) is not true.

But the electric field everywhere may be zero on the surface if there is no charge inside. Option (b) is true.   

Q#5

An electric dipole is placed at the center of a sphere. Mark the correct options;

(a) The flux of the electric field through the sphere is zero.

(b) The electric field is zero at every point of the sphere.

(c) The electric field is not zero anywhere on the sphere.

(d) The electric field is zero on a circle on the sphere.           

Answer: (a), (c). 

Since the total charge of a dipole is zero hence the flux of net electric field through the sphere is zero. Option (a) is correct. 

The electric field at a distance r (radius of the sphere) from the center of a dipole, 

E = $\frac{1}{4\pi \epsilon_0}\frac{p}{r^3}\sqrt{3cos^2 \theta+1}$ 

Where theta is the angle between the radius and the axis of the dipole.

It shows that the electric field is not zero at every point of the sphere. Option (b) is not true.    

If we equate the above value of E to zero, then 3

 cos²θ +1 =0

cos²θ = $-\frac{1}{3}$

which gives cosθ value as imaginary. So the electric field is not zero at any point on the sphere. Option (c) is correct and (d) is not true.    

Q#6    

Figure (30-Q5) shows a charge q placed at the center of a hemisphere. A second charge Q is placed at one of the positions A, B, C, and D. In which position(s) of this second charge, the flux of the electric field through the hemisphere remains unchanged?

(a) A

(b) B

(c) C

(d) D.              

Answer: (a), (c). 

Suppose the area of the surface of the hemisphere = $\Delta S$. The solid angle subtended by this area at a point = $\Delta \Omega$.

Now the flux of the electric field through ΔS due to the charge at any point is given as,

$\Delta \phi = \frac{Q}{4 \pi\epsilon_0}\Delta \Omega$

The Figure for Q#6

To get the solid angle subtended by a curved surface at a point, we join the ends to the concerned point. As we see in the above figure the solid angle Δ𝛀 has some non-zero values at the points B and D but is zero at the points A and C. So the flux of the electric field through the hemisphere by the second charge Q placed at A or C is zero and the original flux due to the charge q at the center remains unchanged. Thus only options (a) and (c) are true. 

Q#7

A closed surface S is constructed around a conducting wire connected to a battery and a switch (figure 30-Q6). As the switch is closed, the free electrons in the wire start moving along the wire. In any time interval, the number of electrons entering the closed surface S is equal to the number of electrons leaving it. On closing the switch, the flux of the electric field through the closed surface

(a) is increased

(b) is decreased

(c) remains unchanged

(d) remains zero.          

Answer: (c), (d). 

A battery is analogous to a water pump connected to a pipeline. The pump creates pressure to make the water already in the pipe to move but does not increase the amount of water in a section of the pipe at any instant. Similarly, the battery makes the pressure to move the free or lose electrons in the metal wire. At any section, the number of electrons is not increased. Hence the wire is not negatively charged. Thus on closing the switch, the charge enclosed by the closed surface S remains zero. So the flux of the electric field through the closed surface remains unchanged and zero. The options (c) and (d) are true only.      

Q#8

Figure (30-Q7) shows a closed surface that intersects a conducting sphere. If a positive charge is placed at point P, the flux of the electric field through the closed surface

(a) will remain zero

(b) will become positive

(c) will become negative

(d) will become undefined.           

Answer: (b) 

The electric field at the sphere due to the positive charge at P will be more or less from right to left. Since the sphere is conducting the loose electrons in it will rearrange themselves to make the electric field at the interior of the sphere to zero. To do this the direction of the electric field due to these rearranged electrons inside the sphere should be equal and opposite to the electric field due to the positive charge at P i.e. from left to right. This can only be when loose electrons on the sphere come towards the sphere surface facing the point P and this face becomes negatively charged. Since the sphere as a whole is uncharged, the face of the sphere away from the point P becomes positively charged. 

Fig. For Q#8

The part of the surface of the sphere inside the closed surface is positively charged (see figure above). Hence the flux through the closed surface will become positive. Option (b) is only correct.