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

Q#1

A charge Q is uniformly distributed over a large plastic plate. The electric field at a point P close to the center of the plate is 10 V/m. If the plastic plate is replaced by a copper plate of the same geometrical dimensions and carrying the same charge Q, the electric field at the point P will become

(a) zero

(b) 5 V/m

(c) 10 V/m

(d) 20 V/m

Answer: (c)

There will be no change. Since the charge is uniformly distributed in both cases, the charge enclosed within the cylindrical Gaussian surface will also be the same. So the electric field will also be the same.

Q#2

A metallic particle having no charge is placed near a finite metal plate carrying a positive charge. The electric force on the particle will be

(a) towards the plate

(b) away from the plate

(c) parallel to the plate

(d) zero

Answer: (a)

Since the
particle is metallic, there are some loose electrons in it. Due to the positive
charge on the nearby plate, the lose electrons in the particle come towards the
plate and there is a deficiency of the electrons on the farther side. Though
the particle as a whole is neutral it is electrically polarized. The negative
charge is near the plate and equal amount of positive charge is away from the
plate. Thus there is an attractive force between the plate and the negative
charge on the particle and a repulsive force between the plate and the positive
charge on the particle. Though the amount of charges involved in both the
forces are the same the distances between the charges are not the same. The
positive charge on the metal particle is farther than the negative charge from
the plate. Since the electric force is inversely proportional to the square of
the distance, the attractive force is stronger than the repulsive force. So the
net force is attractive and towards the plate.

Q#3

A thin metallic spherical shell contains a charge Q on it. A point charge q is placed at the center of the shell and another charge q₁ is placed outside it as shown in figure (30-Q1). All three charges are positive. The force on the charge at the center is

(a) towards left

(b) towards the right

(c) upward

(d) zero.

Answer: (d)

If we know the field at the center of of the spherical metallic shell then the force on the charge q can be known. The field due to the charge Q at the center will be zero. The field inside the metallic shell due to the charge q₁ will be zero because the loose electrons in the metal rearrange themselves and make the field inside zero due to the outer electric field. So the net electric field at the center will be zero. Thus force on the charge q will be zero.

Q#4

Consider the situation of the previous problem. The force on the central charge due to the shell is

(a) towards left

(b) towards the right

(c) upward

(d) zero.

Answer: (b)

In the steady-state, the loose electrons on the metallic shell rearrange such that they concentrate towards the charge q₁ and the opposite side gets positively charge. Thus the net force on the central positive charge due to the shell will be towards the right. (Mind it, the net force on the central charge due to all charges is zero)

Q#5

Electric charges are distributed in a small volume. The flux of the electric field through a spherical surface of radius 10 cm surrounding the total charge is 25 V-m. The flux over a concentric sphere of radius 20 cm will be

(a) 25 V-m

(b) 50 V-m

(c) 100 V-m

(d) 200 V-m

Answer: (a)

The Gauss's Law says, the flux of the net electric field through a closed surface equals the net charge enclosed by the surface divided by εₒ. εₒ is a constant and the net charge enclosed by spherical surfaces of radii 10 cm and 20 cm is the same. Hence, in both cases, the flux is the same $\frac{q}{\epsilon_0}=25 \ V.m$.

Q#6

Figure (30-Q2a) shows an imaginary cube of edge L/2. A uniformly charged rod of length L moves towards the left at a small but constant speed v. At t =0, the left end just touches the center of the face of the cube opposite it. Which of the graphs shown in figure (30-Q2b) represents the flux of the electric field through the cube as the rod goes through it?

Answer: (d)

Suppose the total charge on the rod = q.

Charge per unit length = $\frac{q}{L}$

Length of the rod inside the cube in time t =vt.

Charge inside the cube at time

t = $\frac{q}{L}(vt) = \frac{qv}{L}t$

with a maximum limit of $\frac{q}{2}$.

This limit of $\frac{q}{2}$ will start at time $\frac{L}{2v}$ when half of the rod just enters the cube. And this limit $\frac{q}{2}$ will remain the same until time $\frac{L}{v}$ when the end of the rod enters the cube.

After this withdrawal of the rod begins and the charge remaining inside is

$\frac{3q}{2}-\left(\frac{qv}{L}t\right)$.

The flux of the electric field through the cube = $\frac{Q}{\epsilon_0}$

Between time t = 0 to t = $\frac{L}{2v}$,

*Flux* = $\left(\frac{qv}{L\epsilon_0}\right)t$

So the flux is directly proportional to the time, the graph will be a straight line starting from the origin and slope

$= \frac{qv}{L\epsilon_0}$

Between time t = $\frac{L}{2v}$ to t = $\frac{L}{v}$, the flux is = $\frac{q}{2\epsilon_0}$ = constant. So the graph will be a straight line parallel to the time axis.

Between
time t = $\frac{L}{v}$ to t = $\frac{3L}{2v}$, the flux will be

= $\frac{3q}{2\epsilon_0}-\left(\frac{qv}{L\epsilon_0} \right)t$.

It is also a straight line graph with slope = $-\frac{qv}{L\epsilon_0}$.

The graphs, a, b and c shown in the figure are all straight line. Only the graph d fulfills our explanations. Hence the option (d).

Q#7

A charge q is placed at the center of the open end of a cylindrical vessel (figure 30-Q3). The flux of the electric field through the surface of the vessel is

(a) zero

(b) $\frac{q}{\epsilon_0}$

(c) $\frac{q}{2\epsilon_0}$

(d) $\frac{2q}{\epsilon_0}$

Answer (c)

Assume an
identical inverted cylindrical vessel over the given one.

Now take both combined as a Gaussian surface which is a closed cylindrical vessel. Now the charge q is inside it just at the center. The flux through the whole closed cylindrical surface = $\frac{q}{\epsilon_0}$.

Hence from symmetry, we can say that the flux of the electric field through the surface of the given open vessel = $\frac{q}{2\epsilon_0}$.

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