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#### Page No 139:

#### Answer:

The flux of an electric field $\overrightarrow{E}$ through a surface area $\u2206\overrightarrow{S}$ is given by $\u2206\varphi =\overrightarrow{E}.\u2206\overrightarrow{S}$ , where $\u2206\varphi $ is the flux. Therefore, $\u2206\varphi =E\u2206SCos\theta $. Here, $\theta $ is the angle between the electric field $\overrightarrow{E}$ and the normal to the surface area.

Thus, for the flux to be maximum, cos $\theta $ should be maximum. Thus, for $\theta $ = 0, the flux is maximum, i.e. the electric field lines are perpendicular to the surface area.

The flux is minimum if $\theta $ = 90. Thus, cos $\theta $ = 0 and, hence, flux is also 0. Thus, if the electric field lines are parallel to the surface area, the flux is minimum.

#### Page No 139:

#### Answer:

It is given that the circular ring, made of a non-conducting material, of radius *r *is placed with its axis parallel to a uniform electric field.This means that both the electric field and the area vector are parallel to each other (area vector is always perpendicular to the surface area). Thus, the flux through the ring is given by $\overrightarrow{E}.\overrightarrow{S}$ = *ES *cos 0 = *E*(*$\mathrm{\pi}{r}^{\mathit{2}}$*)*.*

Now, when the ring is rotated about its diameter through 180^{0}, the angle between the area vector and the electric field becomes 180^{0}. Thus, the flux becomes* $-$E $\left(\mathrm{\pi}{r}^{\mathit{2}}\right)$*.

#### Page No 139:

#### Answer:

The field at the centre of the shell is zero. As all the charge given to a conductor resides on the surface, the field at any point inside the conducting sphere is zero. Also, the charge distribution at the surface is uniform; so all the electric field vectors due to these charges at the centre are equal and opposite. So, they cancel each other, resulting in a zero net value of the field.

When a charge is brought near the shell, due to induction, opposite polarity charges induce on the surface nearer to the charge and the same polarity charges appear on the face farther from the charge. In this way, a field is generated inside the shell. Hence, the field at the centre is non-zero.

Yes, our answer changes in case of a non-conducting spherical shell. As the charge given to the surface of a non-conducting spherical shell spreads non-uniformly, there is a net electric field at the centre of the sphere.

#### Page No 139:

#### Answer:

As the shell is made of plastic, it is non-conducting. But as the charge is distributed uniformly over the surface of the shell, the sum of all the electric field vectors at the centre due to this kind of distribution is zero. But when the plastic shell is deformed, the distribution of charge on it becomes non-uniform. In other words, the sum of all the electric field vectors is non-zero now or the electric field exists at the centre now.

In case of a deformed conductor, the field inside is always zero.

#### Page No 139:

#### Answer:

Yes, the charge *Q *will feel an electric force, as the charge q given to the metal block appears on the surface. Hence, it exerts an electric force on the charge *Q*.

#### Page No 139:

#### Answer:

No, the field is not zero everywhere, as the electric field vector due to the charge distribution does not cancel out at any place inside the balloon because of its non-spherical shape.

#### Page No 139:

#### Answer:

Protons never take part in any electrical phenomena because they are inside the nuclei and are not able to interact easily. These are the free electrons that are responsible for all electrical phenomena. So, if a conductor is given a negative charge, the free electrons come to the surface of the conductor. If the conductor is given a positive charge, electrons move away from the surface and leave a positive charge on the surface of the conductor.

#### Page No 139:

#### Answer:

(c) 10 V m^{-1}

The electric field remains same for the plastic plate and the copper plate, as both are considered to be infinite plane sheets. So, it does not matter whether the plate is conducting or non-conducting.

The electric field due to both the plates,

*E* = $\frac{\sigma}{{\epsilon}_{0}}$

#### Page No 139:

#### Answer:

(a) towards the plate

The particle is a conductor. When it is brought near a positively charged metal plate, opposite charge is induced on its face nearer to the plate, i.e. negative and the same amount of charge, but of opposite polarity, goes to the farther end, i.e. positive.

Now, the attractive force is due to charges of opposite polarity. As they are at a lesser distance than the same polarity charges, the force of attraction is greater than the force of repulsion. In other words, the force on the particle is towards the plate.

#### Page No 139:

#### Answer:

(d) zero

A charge placed outside a conductor can induce a charge on it or can affect the charge on its surface. But it does not affect what is contained inside the conductor. Similarly, charge *q*_{1} can affect charge Q; still, the force inside the conductor remains zero. An analogy to the above statement is that when lightning strikes a car, the charge that appears on the car's metal surface does not affect its interior. Due to this passengers are recommended to sit inside the car.

#### Page No 140:

#### Answer:

(b) towards right

This question can be answered using the concept of electric field lines.

We know that electric field lines emerge from a positive charge and move towards a negative charge. Now, charge *Q*, on the face in front of charge *q*_{1}_{,}_{ }tries to nullify the field lines emerging from charge *q*_{1}_{.} So, the charge on the farther face of the conductor dominates and, hence, a force appears to the right of charge *q*.

#### Page No 140:

#### Answer:

(a) 25 V m

The flux through a surface does not depend on its shape and size; it only depends upon the charge enclosed inside the volume. Here, the charge enclosed by the sphere of radius 10 cm and the sphere of radius 20 cm is same so the flux through them will also be same.

#### Page No 140:

#### Answer:

(d)

At first, when the rod is inserted into the cube, the flux start increasing. When the rod is fully inserted, the flux becomes constant and remains constant for the remaining *L*/2 length of the rod. After that, as the rod moves out of the cube, the flux starts decreasing. These processes are depicted only by curve (*d*).

#### Page No 140:

#### Answer:

(C) *q*/2${\in}_{0}$

According to Gauss's Law, the flux through a closed cylindrical Gaussian surface is *q*/${\in}_{0}$. But the question is about an open cylindrical vessel. Now, take another identical vessel and make a closed Gaussian surface enclosing the charge, as shown in the figure.

Total flux linked with the closed Gaussian surface,

${\varphi}_{T}=\frac{q}{{\in}_{0}}$

Flux linked with the surface of a open ended cylindrical vessel,

$\varphi =\frac{{\varphi}_{T}}{2}=\frac{q}{2{\in}_{0}}$

#### Page No 140:

#### Answer:

(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.

The contribution of flux on the closed surface due to the charges lying outside the surface is zero because number of field line entering the closed surface is equal to the number of field lines coming out of the surface so the net contribution of the charge lying outside the closed surface to the flux is zero. Therefore, the net flux through the surface due to the charge lying outside the the closed surface is zero. The contribution that counts is only due to the charges lying within the closed surface.

Thus, the flux of the electric field through a closed surface due to all the charges (inside and outside the surface) is equal to the flux due to the charges enclosed by the surface.

#### Page No 140:

#### Answer:

(d) The interior remains charge free and the surface gets non-uniform charge distribution.

Since the cube is metallic, the charge gets distributed on the surface and the interior remains charge-free. However, when a positive point charge *Q *is brought near the metallic cube, a negative charge gets induced on the face near the positive charge *Q *and an equal positive charge gets induced on the face, which is away from the charge *Q*. Thus, the metallic surface gets non-uniform charge distribution.

#### Page No 140:

#### Answer:

*M*attracts

*A*.

(b)

*M*attracts

*B*.

(c)

*A*attracts

*B*.

(d)

*B*attracts

*A*.

Since the non-conducting sheet

*M*is given a uniform charge, it induces a charge in the metal rod

*A,*which further induces a charge in the metal rod

*B,*as shown in the figure. Hence, all the options are correct.

#### Page No 140:

#### Answer:

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

(c) the charge inside the surface must be zero

As the flux is zero through the surface, the charge enclosed must be zero. But the electric field is not necessarily zero everywhere on the surface. For example, in the case of a dipole enclosed by the surface, the electric field through the surface is not zero but has some value. So, the correct answers are (b) and (c).

#### Page No 140:

#### Answer:

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

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

The sphere encloses a dipole, i.e. two equal and opposite charges. In other words, net charge enclosed in the sphere is zero. Hence, the flux is zero through the sphere.

But the electric field at any point *p* on the sphere,

$E=\frac{1}{4\pi {\in}_{0}}\frac{p}{{r}^{3}}\sqrt{3{\mathrm{cos}}^{2}\theta +1}$ , where $\theta $ is the angle made by the point *p* with the centre of the dipole.

Hence, we can see that the field is not zero anywhere on the sphere.

#### Page No 140:

#### Answer:

(a) *A*

(c) C

These are the only points in the straight line with the charge *q* and at the brim of the hemisphere. So, field lines emerging from these charges do not affect the flux through the hemisphere due to charge* q*. On the other hand, the two remaining charges are beside the surface of the hemisphere and not in line with the charge *q* and not at the brim. So, they will affect the flux. Hence, the correct answers are (a) and (c).

#### Page No 140:

#### Answer:

(c) remains unchanged

(d) remains zero

Initially, there is no charge in the closed surface. As the wire is neutral, the flux initially is zero. Now, if we connect the battery and a current flows through it, the flux remains zero, as the number of electrons entering the surface is equal to number of electrons leaving. That is, net charge enclosed is zero and so is the flux.

#### Page No 141:

#### Answer:

(b) will become positive

A positive charge at point P will induce a negative charge on the near face of the conducting sphere, whereas the positive charge on the farther end of the sphere. As this father end is enclosed or intersected by the closed surface, so the flux through it will become positive due to the induced positive charge.

#### Page No 141:

#### Answer:

Given:

Electric field strength, $\overrightarrow{\mathit{E}}=\frac{3}{5}{E}_{0}\hat{\mathit{i}}+\frac{4}{5}{E}_{0}\stackrel{\u23dc}{j}$,

where ${E}_{0}=2.0\times {10}^{3}\mathrm{N}/\mathrm{C}$

The plane of the rectangular surface is parallel to the *y-z* plane. The normal to the plane of the rectangular surface is along the *x* axis. Only $\frac{3}{5}{E}_{0}\stackrel{\u23dc}{i}$ passes perpendicular to the plane; so, only this component of the field will contribute to flux.

On the other hand, $\frac{4}{5}{E}_{0}\stackrel{\u23dc}{j}$ moves parallel to the surface.

Surface area of the rectangular surface, *a *= 0⋅2 m^{2}

Flux,

$\varphi =\overrightarrow{E}.\overrightarrow{a}=E\times a\phantom{\rule{0ex}{0ex}}\varphi =\left(\frac{3}{5}\times 2\times {10}^{3}\right)\times \left(2\times {10}^{-1}\right){\mathrm{Nm}}^{2}/\mathrm{C}\phantom{\rule{0ex}{0ex}}\varphi =0.24\times {10}^{3}{\mathrm{Nm}}^{2}/\mathrm{C}\phantom{\rule{0ex}{0ex}}\varphi =240{\mathrm{Nm}}^{2}/\mathrm{C}$

#### Page No 141:

#### Answer:

Given:

Total charge on the rod = *Q*

The length of the rod = edge of the hypothetical cube =* l*

Portion of the rod lying inside the cube, *x* = *$\frac{l}{2}$*

Linear charge density for the rod *$=\frac{Q}{l}$*

Using Gauss's theorem, flux through the hypothetical cube,

*ϕ* =(*Q*_{in}/${\in}_{0}$), where *Q*_{in}_{ }= charge enclosed inside the cube

Here, charge per unit length of the rod = $\frac{Q}{l}$

Charge enclosed, *Q*_{in} = $\frac{Q}{l}\times \frac{l}{2}=\frac{Q}{2}$

Therefore,

*ϕ*$=\frac{Q/2}{{\mathit{\in}}_{0}}=\frac{Q}{2{\in}_{0}}$

#### Page No 141:

#### Answer:

It is given that the electric field is uniform. If we consider a surface perpendicular to the electric field, we find that it is an equipotential surface. Hence, if a test charge is introduced on the surface, then work done will be zero in moving the test charge on it.

But if there is some net charge in this region, the test charge introduced on the surface will experience a force due to this charge. This force has a component parallel to the surface; thus, work has to be done in moving this test charge. Thus, the surface cannot be said to be equipotential. This implies that the net charge in the region with uniform electric field is zero.

#### Page No 141:

#### Answer:

Given:

Electric field strength, $\overrightarrow{E}=\frac{{E}_{0}x}{l}\hat{i}$

Length, $l=2\mathrm{cm}$

Edge of the cube, $a=1\mathrm{cm}$

${E}_{0}=5.0\times {10}^{3}\mathrm{N}/\mathrm{C}$

It is observed that the flux passes mainly through the surfaces ABCD and EFGH. The surfaces AEFB and CHGD are parallel to the electric field. So, electric flux for these surfaces is zero.

The electric field intensity at the surface EFGH will be zero.

If the charge is inside the cube, then equal flux will pass through the two parallel surfaces ABCD and EFGH. We can calculate flux passing only through one surface. Thus,

$\overrightarrow{E}=\frac{{E}_{0}x}{l}\hat{i}$

At EFGH, *x* = 0; thus, the electric field at EFGH is zero.

At ABCD, *x* = *a;* thus, the electric field at ABCD is $\overrightarrow{E}=\frac{{E}_{0}a}{l}\hat{i}$.

The net flux through the whole cube is only through the side ABCD and is given by $\varphi =\overrightarrow{E}.\overrightarrow{A}=\left(\frac{{E}_{0}a}{l}\hat{i}\right).\left(a\hat{i}\right)=\frac{{E}_{0}{a}^{2}}{l}$.

Net flux = $\varphi =\frac{5\times {10}^{3}}{2\times {10}^{-2}}\times {\left(1\times {10}^{-2}\right)}^{2}{\mathrm{Nm}}^{2}/\mathrm{C}=25{\mathrm{Nm}}^{2}/\mathrm{C}$

Thus, the net charge,

$q={\in}_{0}\varphi \phantom{\rule{0ex}{0ex}}q=8.85\times {10}^{-12}\times 25\phantom{\rule{0ex}{0ex}}q=22.125\times {10}^{-13}\phantom{\rule{0ex}{0ex}}q=2.2125\times {10}^{-12}\mathrm{C}$

#### Page No 141:

#### Answer:

According to Gauss's Law, flux passing through any closed surface is equal to $\frac{1}{{\in}_{0}}$ times the charge enclosed by that surface.

$\Rightarrow \varphi =\frac{q}{{\mathit{\in}}_{0}}$,

where *ϕ* is the flux through the closed surface and *q* is the charge enclosed by that surface.

The charge is placed at the centre of the cube and the electric field is passing through the six surfaces of the cube.vSo, we can say that the total electric flux passes equally through these six surfaces .

Thus, flux through each surface,

$\varphi \text{'}=\frac{Q}{6{\in}_{0}}$

#### Page No 141:

#### Answer:

Given:

Edge length of the square surface = *a*

Distance of the charge Q from the square surface = *a*/2

Area of the plane = *a*^{2}

Assume that the given surface is one of the faces of the imaginary cube.

Then, the charge is found to be at the centre of the cube.

A charge is placed at a distance of about $\frac{a}{2}$ from the centre of the surface.

The electric field due to this charge is passing through the six surfaces of the cube.

Hence flux through each surface,

$\varphi =\frac{Q}{{\mathit{\in}}_{\mathit{0}}}\mathit{\times}\frac{1}{6}\mathit{=}\frac{Q}{\mathit{6}\mathit{}{\mathit{\in}}_{\mathit{0}}}$

Thus, the flux through the given surface is $\frac{Q}{6{\in}_{0}}$.

#### Page No 141:

#### Answer:

Given:

Let charge *Q* be placed at the centre of the sphere and *Q*' be placed at a distance 2R from the centre.

Magnitude of the two charges = 10^{−7} C

According to Gauss's Law, the net flux through the given sphere is only due to charge *Q* that is enclosed by it and not by the charge *Q*' that is lying outside.

So, only the charge located inside the sphere will contribute to the flux passing through the sphere.

Thus,

$\varphi =\int \overrightarrow{E}.d\overrightarrow{s}\mathit{=}\frac{Q}{{\mathit{\in}}_{\mathit{0}}}=\frac{{10}^{-7}}{8.85\times {10}^{-12}}\phantom{\rule{0ex}{0ex}}\Rightarrow \varphi =1.1\times {10}^{4}{\mathrm{Nm}}^{2}{\mathrm{C}}^{-1}$

#### Page No 141:

#### Answer:

From Guass's law, flux through a closed surface,

$\varphi =\frac{{Q}_{\mathrm{en}}}{{\in}_{0}}$,

where

*Q*_{en} = charge enclosed by the closed surface

Let us assume that a spherical closed surface in which the charge is enclosed is *Q*.

The flux through the sphere,

$\varphi =\frac{Q}{{\mathit{\in}}_{0}}$

Hence for a hemisphere(open bowl), total flux through its curved surface,

$\varphi \text{'}=\frac{Q}{{\in}_{0}}\times \frac{1}{2}=\frac{Q}{2{\in}_{0}}$

#### Page No 141:

#### Answer:

Given :

Volume charge density, *ρ* = $2\times {10}^{-4}\mathrm{C}/{\mathrm{m}}^{3}$

Let us assume a concentric spherical surface inside the given sphere with radius =$4\mathrm{cm}=4\times {10}^{-2}\mathrm{m}$

The charge enclosed in the spherical surface assumed can be found by multiplying the volume charge density with the volume of the sphere. Thus,

$q=\rho \times \frac{4}{3}\mathrm{\pi}{r}^{3}\phantom{\rule{0ex}{0ex}}\Rightarrow q=\left(2\times {10}^{-4}\right)\times \frac{4}{3}\mathrm{\pi}{r}^{3}\phantom{\rule{0ex}{0ex}}$

The net flux through the spherical surface,

$\varphi =\frac{q}{{\in}_{0}}$

The surface area of the spherical surface of radius *r* cm:

*A* = $4\pi {r}^{2}$

Electric field,

$E=\frac{q}{{\in}_{0}\times A}\phantom{\rule{0ex}{0ex}}E=\frac{2\times {10}^{-4}\times 4\mathrm{\pi}{r}^{3}}{{\in}_{0}\times 3\times 4\mathrm{\pi}{r}^{2}}\phantom{\rule{0ex}{0ex}}E=\frac{2\times {10}^{-4}\times r}{3\times {\in}_{0}}$

The electric field at the point inside the volume at a distance 4⋅0 cm from the centre,

$E=\frac{(2\times {10}^{-4})\times (4\times {10}^{-2})}{3\times (8.85\times {10}^{-12})}\mathrm{N}/\mathrm{C}\phantom{\rule{0ex}{0ex}}E=3.0\times {10}^{5}\mathrm{N}/\mathrm{C}$

#### Page No 141:

#### Answer:

Given:

Atomic number of gold = 79

Charge on the gold nucleus, $Q=79\times (1.6\times {10}^{-19})\mathrm{C}$

The charge is distributed across the entire volume. So, using Gauss's Law, we get:

(a)

$\varphi =\oint \overrightarrow{\mathrm{E}}.d\overrightarrow{s}=\frac{Q}{{\in}_{0}}\phantom{\rule{0ex}{0ex}}\Rightarrow \oint Eds=\frac{Q}{{\in}_{0}}\phantom{\rule{0ex}{0ex}}$

The value of E is fixed for a particular radius.

$\Rightarrow E\oint ds=\frac{Q}{{\in}_{0}}\phantom{\rule{0ex}{0ex}}\Rightarrow E\times 4\pi {r}^{2}=\frac{Q}{{\in}_{0}}\phantom{\rule{0ex}{0ex}}\Rightarrow E=\frac{Q}{{\in}_{0}\times 4\pi {r}^{2}}\phantom{\rule{0ex}{0ex}}$

$E=\frac{79\times (1.6\times {10}^{-19})}{(8.85\times {10}^{-12})\times 4\times 3.14\times (7\times {10}^{-10}{)}^{2}}\phantom{\rule{0ex}{0ex}}E=2.315131\times {10}^{21}\mathrm{N}/\mathrm{C}$

(b) To find the electric field at the middle point of the radius:

Radius, $r=\frac{7}{2}\times {10}^{-10}\mathrm{m}$

Volume,

$V=\frac{4}{3}\mathrm{\pi}{r}^{3}\phantom{\rule{0ex}{0ex}}=\frac{4}{3}\times \frac{22}{7}\times \frac{343}{8}\times {10}^{-30}$

Net charge $=79\times 1.6\times {10}^{-19}\mathrm{C}$

Volume charge density

$=\frac{79\times 1.6\times {10}^{-19}}{{\displaystyle \frac{4}{3}}\mathrm{\pi}\times 343\times {10}^{-30}}$

So, the charge enclosed by this imaginary sphere of radius *r* =3.5 $\times {10}^{-10}$

$=\frac{79\times 1.6\times {10}^{-19}}{{\displaystyle \frac{4}{3}}\mathrm{\pi}\times 343\times {10}^{-30}}\times \frac{4}{3}\mathrm{\pi}\times \frac{343}{8}\times {10}^{-30}\phantom{\rule{0ex}{0ex}}=\frac{79\times 1.6\times {10}^{-19}}{8}$

$\Rightarrow E=\frac{79\times 1.6\times {10}^{-19}}{8\times 4\mathrm{\pi}{\in}_{0}.{\mathrm{r}}^{2}}\mathrm{at}r\mathit{}=3.5\times {10}^{-10}\phantom{\rule{0ex}{0ex}}=1.16\times {10}^{21}\mathrm{N}/\mathrm{C}.$

As electric charge is given to a conductor, it gets distributed on its surface. But nucleons are bound by the strong force inside the nucleus. Thus, the nuclear charge does not come out and reside on the surface of the conductor. Thus, the charge can be assumed to be uniformly distributed in the entire volume of the nucleus.

#### Page No 141:

#### Answer:

Amount of charge present on the hollow sphere = *Q*

Inner radii of the hollow sphere = *${r}_{1}$*

Outer radii of the hollow sphere = ${r}_{2}$

Consider an imaginary sphere of radius *x*.

The charge on the sphere can be found by multiplying the volume charge density of the hollow spherical volume with the volume of the imaginary sphere of radius (*x*-${r}_{1}$).

Charge per unit volume of the hollow sphere,

$\rho =\frac{Q}{{\displaystyle \frac{4}{3}}\pi \left({r}_{2}^{3}-{r}_{1}^{3}\right)}$

Charge enclosed by this imaginary sphere of radius *x:*

*q* = *ρ *× Volume of the part consisting of charge

$q=\frac{{\displaystyle \frac{4}{3}}\mathrm{\pi}\left({x}^{3}-{r}_{1}^{3}\right)Q}{{\displaystyle \frac{4}{3}}\mathrm{\pi}\left({r}_{2}^{3}-{r}_{1}^{3}\right)}\phantom{\rule{0ex}{0ex}}q=\frac{\left({x}^{3}-{r}_{1}^{3}\right)}{\left({r}_{2}^{3}-{r}_{1}^{3}\right)}Q$

According to Gauss's Law,

$\oint \mathrm{E}.ds=\frac{q}{{\in}_{0}}$

Here, the surface integral is carried out on the sphere of radius *x* and *q* is the charge enclosed by this sphere.

$\mathrm{E}\oint ds=\frac{q}{{\in}_{0}}\phantom{\rule{0ex}{0ex}}E\left(4\mathrm{\pi}{x}^{2}\right)=\frac{q}{{\in}_{0}}$

$E(4\mathrm{\pi}{x}^{2})=\frac{\left({x}^{3}-{r}_{1}^{3}\right)Q}{\left({r}_{2}^{3}-{r}_{1}^{3}\right){\mathit{\in}}_{0}}\phantom{\rule{0ex}{0ex}}E\mathit{}=\frac{Q\left({x}^{3}-{r}_{1}^{3}\right)}{4\mathrm{\pi}{\mathit{\in}}_{0}{\mathrm{x}}^{2}\left({\mathrm{r}}_{2}^{3}-{\mathrm{r}}_{1}^{3}\right)}$

The electric field is directly proportional to x for *r*_{1} < *x* < *r*.

The electric field for *r*_{2} < *x* < 2*r*_{2,}

$E=\frac{Q}{4\pi {\in}_{0}{x}^{2}}$

Thus, the graph can be drawn as:

#### Page No 142:

#### Answer:

Given:

Amount of charge present at the centre of the hollow sphere = *Q*

We know that charge given to a hollow sphere will move to its surface.

Due to induction, the charge induced at the inner surface = −*Q*

Thus, the charge induced on the outer surface = +*Q*

(a)

Surface charge density is the charge per unit area, i.e.

*σ*$=\frac{\mathrm{charge}}{\mathrm{total}\mathrm{surface}\mathrm{area}}$

Surface charge density of the inner surface, ${\sigma}_{\mathrm{in}}=\frac{-Q}{4\mathrm{\pi}{a}^{2}}$

Surface charge density of the outer surface, ${\sigma}_{\mathrm{out}}=\frac{Q}{4\mathrm{\pi}{a}^{2}}$

(b)

Now if another charge *q* is added to the outer surface, all the charge on the metal surface will move to the outer surface. Thus, it will not affect the charge induced on the inner surface. Hence the inner surface charge density,

${\sigma}_{\mathrm{in}}=-\frac{q}{4\pi {a}^{2}}$

As the charge has been added to the outer surface, the total charge on the outer surface will become (*Q+q*).

So the outer surface charge density, ${\sigma}_{\mathrm{out}}=\frac{q+Q}{4\pi {a}^{2}}$

(c)

To find the electric field inside the sphere at a distance *x* from the centre in both the situations,let us assume an imaginary sphere inside the hollow sphere at a distance x from the centre.

Applying Gauss's Law on the surface of this imaginary sphere,we get:

$\oint E.ds=\frac{Q}{{\mathit{\in}}_{0}}\phantom{\rule{0ex}{0ex}}E\oint ds=\frac{Q}{{\mathit{\in}}_{0}}\phantom{\rule{0ex}{0ex}}E\left(4\mathrm{\pi}{x}^{2}\right)=\frac{Q}{{\mathit{\in}}_{0}}\phantom{\rule{0ex}{0ex}}\mathit{}E=\frac{Q}{{\mathit{\in}}_{0}}\times \frac{1}{4\mathrm{\pi}{x}^{2}}=\frac{Q}{4\mathrm{\pi}{\mathit{\in}}_{0}{x}^{2}}$

Here, *Q* is the charge enclosed by the sphere.

For situation (b):

As the point is inside the sphere, there is no effect of the charge q given to the shell.

Thus, the electric field at the distance x:

$\mathit{}E=\frac{Q}{4\mathrm{\pi}{\mathit{\in}}_{0}{x}^{2}}$

#### Page No 142:

#### Answer:

(a)

Let us consider the three surfaces as three concentric spheres A, B and C.

Let us take *q* = $1.6\times {10}^{-19}\mathrm{C}$ .

Sphere A is the nucleus; so, the charge on sphere A, ${q}_{1}=4q$

Sphere B is the sphere enclosing the nucleus and the 2 1s electrons; so charge on this sphere, ${q}_{2}=4q-2q=2q$

Sphere C is the sphere enclosing the nucleus and the 4 electrons of Be; so, the charge enclosed by this sphere,${q}_{3}=4q-4q=0$

Radius of sphere A, ${r}_{1}={10}^{-15}\mathrm{m}$

Radius of sphere B, ${r}_{2}=1.3\times {10}^{-11}\mathrm{m}$

Radius of sphere C, ${r}_{3}=5.2\times {10}^{-11}\mathrm{m}$

As the point 'P' is just inside the spherical cloud 1s, its distance from the centre

$x=1.3\times {10}^{-11}\mathrm{m}\phantom{\rule{0ex}{0ex}}$

Electric field,

$E=\frac{q}{4\mathrm{\pi}{\mathit{\in}}_{0}{x}^{2}}$

Here, the charge enclosed is due to the charge of the 4 protons inside the nucleus. So,

$E=\frac{4\times \left(1.6\times {10}^{-19}\right)}{4\times 3.14\times \left(8.85\times {10}^{-12}\right)\times {\left(1.3\times {10}^{-11}\right)}^{2}}\phantom{\rule{0ex}{0ex}}E=3.4\times {10}^{13}\mathrm{N}/\mathrm{C}$

(b)

For a point just inside the 2s cloud, the total charge enclosed will be due to the 4 protons and 2 electrons. Charge enclosed,

${q}_{\mathrm{en}}=2q=2\times \left(1.6\times {10}^{-19}\right)\mathrm{C}$

Hence, electric field,

$E=\frac{{q}_{\mathrm{en}}}{4\mathrm{\pi}{\mathit{\in}}_{0}{x}^{2}}$

*x *= 5⋅2 × 10^{−11} m

$E=\frac{2\times \left(1.6\times {10}^{-19}\right)}{4\times 3.14\times \left(8.85\times {10}^{-12}\right)\times {\left(5.2\times {10}^{-11}\right)}^{2}}\phantom{\rule{0ex}{0ex}}E=1.065\times {10}^{12}\mathrm{N}/\mathrm{C}$

Thus, $E=1.1\times {10}^{12}\mathrm{N}/\mathrm{C}$

#### Page No 142:

#### Answer:

Given:

Charge density of the line containing charge, *λ* = $2\times {10}^{-6}\mathrm{C}{\mathrm{m}}^{-1}$

We need to find the electric field at a distance of 4 cm away from the line charge.

We take a Gaussian surface around the line charge of cylindrical shape of radius *r* = 4 ×10^{$-$2} m and height *l*.

The charge enclosed by the Gaussian surface, *q*_{en} = *λl*

Let the magnitude of the electric field at a distance of 4 cm away from the line charge be *E*.

Thus, net flux through the Gaussian surface, $\varphi =\oint \overrightarrow{E}.d\overrightarrow{s}$

There will be no flux through the circular bases of the cylinder; there will be flux only from the curved surface.

The electric field lines are directed radially outward, from the line charge to the cylinder.

Therefore, the lines will be perpendicular to the curved surface.

Thus,

$\varphi =\oint \overrightarrow{E}.\mathrm{d}\overrightarrow{s}=E\oint \mathrm{d}s\phantom{\rule{0ex}{0ex}}\Rightarrow \varphi =E\times 2\pi rl$

Applying Gauss's theorem,

$\varphi =\frac{{q}_{\mathrm{en}}}{{\in}_{0}}\phantom{\rule{0ex}{0ex}}E\times 2\pi rl=\frac{\lambda l}{{\in}_{0}}\phantom{\rule{0ex}{0ex}}E=\frac{\lambda}{2\pi r{\in}_{0}}$^{}

$E=\frac{2\times {10}^{-6}}{2\times 3.14\times 8.85\times {10}^{-12}\times 4\times {10}^{-2}}\phantom{\rule{0ex}{0ex}}E=8.99\times {10}^{5}\mathrm{N}/\mathrm{C}$

#### Page No 142:

#### Answer:

Let the linear charge density of the wire be $\lambda $.

The electric field due to a charge distributed on a wire at a perpendicular distance *r* from the wire,

$\mathrm{E}=\frac{\lambda}{2\mathrm{\pi}{\in}_{0}r}$

The electrostatic force on the electron will provide the electron the necessary centripetal force required by it to move in a circular orbit. Thus,

$qE=\frac{m{v}^{2}}{r}\phantom{\rule{0ex}{0ex}}\Rightarrow m{v}^{2}=qEr...(1)$

Kinetic energy of the electron, $K=\frac{1}{2}m{v}^{2}$

From (1),

$K=\frac{qEr}{2}\phantom{\rule{0ex}{0ex}}K=\frac{qr}{2}\frac{\lambda}{2\mathrm{\pi}{\in}_{0}r}\left[\because E=\frac{\lambda}{2\mathrm{\pi}{\in}_{0}r}\right]\phantom{\rule{0ex}{0ex}}K=(1.6\times {10}^{-19})\times (2\times {10}^{-8})\times (9\times {10}^{9})\mathrm{J}\phantom{\rule{0ex}{0ex}}K=2.88\times {10}^{-17}\mathrm{J}$

#### Page No 142:

#### Answer:

Given:

Volume charge density inside the cylinder = $\mathrm{\rho}$

Let the radius of the cylinder be *r.*

Let charge enclosed by the given cylinder be *Q*

Consider a Gaussian cylindrical surface of radius *x* and height *h.*

Let charge enclosed by the cylinder of radius *x* be ${q}^{\text{'}}$.

The charge on this imaginary cylinder can be found by taking the product of the volume charge density of the cylinder and the volume of the imaginary cylinder. Thus,

${q}^{\text{'}}$=$\rho \left(\pi {x}^{\mathit{2}}h\right)$

From Gauss's Law,

$\oint E\mathit{.}ds=\frac{{q}_{\mathrm{en}}}{{\in}_{0}}\phantom{\rule{0ex}{0ex}}E.2\mathrm{\pi xh}=\frac{\rho (\pi {x}^{2}h)}{{\in}_{0}}\phantom{\rule{0ex}{0ex}}\mathrm{E}=\frac{\rho x}{2{\in}_{0}}$

#### Page No 142:

#### Answer:

Given:

Thickness of the sheet = *d*

Let the surface area of the sheet be *s*.

Volume of the sheet = *sd*

Volume charge density of the sheet, $\rho =\frac{Q}{sd}$

Charge on the sheet = *Q*

Consider an imaginary plane at a distance *x* from the central plane of surface area *s.*

Charge enclosed by this sheet, *q* = $\rho sx$

For this Guassian surface, using Gauss's Law,we get:

$\oint E.ds=\frac{q}{{\in}_{0}}\phantom{\rule{0ex}{0ex}}E.s=\frac{\rho sx}{{\in}_{0}}\phantom{\rule{0ex}{0ex}}E=\frac{\rho x}{{\in}_{0}}$

The electric field outside the sheet will be constant and will be:

$E=\frac{\rho d}{{\in}_{0}}$

#### Page No 142:

#### Answer:

The electric field due to a conducting thin sheet,

$E=\frac{\sigma}{2{\in}_{0}}$

The magnitude of attractive force between the particle and the plate,

$F\mathit{=}qE\phantom{\rule{0ex}{0ex}}F\mathit{=}\frac{\mathit{q}\mathit{\times}\mathit{\sigma}}{\mathit{2}{\mathit{\in}}_{\mathit{0}}}\phantom{\rule{0ex}{0ex}}F=\frac{\left(2.0\times {10}^{-6}\right)\times \left(4.0\times {10}^{-6}\right)}{2\times \left(8.85\times {10}^{-12}\right)}\phantom{\rule{0ex}{0ex}}F=0.45\mathrm{N}$

#### Page No 142:

#### Answer:

There are two forces acting on the ball. These are

(1) Weight of the ball, *W = mg*

(2) Coulomb force acting on the charged ball due to the electric field of the plate, *F* = *qE*

Due to these forces,a tension develops in the thread.

Let the surface charge density on the plate be *σ*.

Electric field of a plate,

*E* = $\frac{\sigma}{2{\in}_{0}}$

It is given that in equilibrium, the thread makes an angle of 60° with the vertical.

Resolving the tension in the string along horizontal and vertical directions, we get:

$T\mathrm{cos}60\xb0=mg\phantom{\rule{0ex}{0ex}}T\mathrm{sin}60\xb0=\mathit{}qE\phantom{\rule{0ex}{0ex}}\Rightarrow \mathrm{tan}60\xb0=\frac{\mathit{q}\mathit{E}}{\mathit{m}\mathit{g}}\phantom{\rule{0ex}{0ex}}\Rightarrow E=\frac{m\mathrm{g}\mathrm{tan}60\xb0}{q}$

Also, electric field due to a plate,

$E=\frac{\sigma}{2{\in}_{0}}=\frac{mg\mathrm{tan}{60}^{\mathrm{o}}}{q}\phantom{\rule{0ex}{0ex}}\sigma =\frac{2{\in}_{0}m\mathrm{g}\mathrm{tan}60\xb0}{q}\phantom{\rule{0ex}{0ex}}\sigma =\frac{2\times \left(8.85\times {10}^{-12}\right)\times \left(10\times {10}^{-3}\times 9.8\right)\times 1.732}{4.0\times {10}^{-6}}\phantom{\rule{0ex}{0ex}}\sigma =7.5\times {10}^{-7}\mathrm{C}/{\mathrm{m}}^{2}$

#### Page No 142:

#### Answer:

(a)

In equilibrium state, the thread makes an angle of 60^{o} with the vertical.

The tension in the thread is resolved into horizontal and vertical components.

Then, tension in the string in equilibrium,

$T\mathit{}\mathrm{cos}60\xb0=m\mathrm{g}\phantom{\rule{0ex}{0ex}}T\times \frac{1}{2}=\left(10\times {10}^{-3}\right)\times 10\phantom{\rule{0ex}{0ex}}T=\left(10\times {10}^{-3}\right)\times 10\times 2=0.20\mathrm{N}$

(b) As it is displaced from equilibrium, net force on the ball,

$F=\sqrt{{\left(mg\right)}^{2}+{\left(\frac{q\sigma}{2{\in}_{0}}\right)}^{2}}$

As* F = ma*

$\Rightarrow a=\sqrt{{\left(g\right)}^{2}+{\left(\frac{q\sigma}{m2{\in}_{0}}\right)}^{2}}$

The surface charge density of the plate (as calculated in the previous question), *σ* = 7.5×10^{$-$7} C/m^{2}

Charge on the ball, *q* = 4×10^{$-$6 }C

Mass of the ball, m =

The time period of oscillation of the given simple pendulum,

$T=2\mathrm{\pi}\sqrt{\frac{l}{\mathrm{g}}}\phantom{\rule{0ex}{0ex}}=2\mathrm{\pi}\sqrt{\frac{10\times {10}^{-2}}{9.8}}\phantom{\rule{0ex}{0ex}}=0.45\mathrm{sec}$

#### Page No 142:

#### Answer:

Distance travelled by the electron, *d*= 2 cm

Time taken to cross the region, *t* = 2×10^{$-$6} s

Let the surface charge density at the conducting plates be *σ*.

Let the acceleration of the electron be *a*.

Applying the 2nd equation of motion, we get:

$d=\frac{1}{2}a{t}^{2}\phantom{\rule{0ex}{0ex}}\Rightarrow a=\frac{2d}{{t}^{2}}\phantom{\rule{0ex}{0ex}}\phantom{\rule{0ex}{0ex}}$

This acceleration is provided by the Coulombic force. So,

$a=\frac{qE}{m}=\frac{2d}{{t}^{2}}\phantom{\rule{0ex}{0ex}}\Rightarrow E=\frac{2md}{q{t}^{2}}\phantom{\rule{0ex}{0ex}}E=\frac{2\times (9.1\times {10}^{-31})\times (2\times {10}^{-2})}{(1.6\times {10}^{-19})\times (4\times {10}^{-12})}\phantom{\rule{0ex}{0ex}}E=5.6875\times {10}^{-2}\mathrm{N}/\mathrm{C}$

Also, we know that electric field due to a plate,

$E=\frac{\sigma}{{\in}_{0}}\phantom{\rule{0ex}{0ex}}\Rightarrow \sigma ={\in}_{0}E\phantom{\rule{0ex}{0ex}}\Rightarrow \sigma =\left(8.85\times {10}^{-12}\right)\times \left(5.68\times {10}^{-2}\right)\mathrm{C}/{\mathrm{m}}^{2}\phantom{\rule{0ex}{0ex}}\Rightarrow \sigma =50.33\times {10}^{-14}\mathrm{C}/{\mathrm{m}}^{2}=0.503\times {10}^{-12}\mathrm{C}/{\mathrm{m}}^{2}$

#### Page No 142:

#### Answer:

Given :

Surface charge density on the plates $=\sigma $

The electric field due to plate 1, *E*_{1} = $\frac{\sigma}{2{\in}_{0}}$

The electric field due to plate 2, *E*_{2} = $\frac{\sigma}{2{\in}_{0}}$

(a) The strength of the electric field due to both the plates will be same but their directions will be opposite to each other on any point at the left of the two plates.

Thus, the net electric field at a point on the left of plate1 = $\frac{\sigma}{2{\in}_{0}}-\frac{\sigma}{2{\in}_{0}}=0$

(b) Here the direction of the fields will be same. So, they will add up to give the resultant field in this region.

Total electric field:

$\frac{\sigma}{2{\in}_{0}}+\frac{\sigma}{2{\in}_{0}}=\frac{\sigma}{{\in}_{0}}$

(c) The strength of the electric field due to both the plates will be same but their directions will be opposite to each other at any point on the right of the two plates.

Thus, the net electric field at a point on the left of plate 2 = $\frac{\sigma}{2{\in}_{0}}-\frac{\sigma}{2{\in}_{0}}=0$

#### Page No 142:

#### Answer:

(a)

Given that the charge present on the plate is *Q.* The other plate will get the same charge* Q *due* *to convection*.*

Let the surface charge densities on both sides of the plate be *σ*_{1} and *σ*_{2}.

Now, electric field due to a plate,

$E=\frac{\sigma}{2{\in}_{0}}$

So, the magnitudes of the electric fields due to this plate on each side

$=\frac{{\sigma}_{1}}{2{\in}_{0}}\mathrm{and}\frac{{\sigma}_{2}}{2{\in}_{0}}$

The plate has two sides, each of area *A. *So, the net charge given to the plate will be equally distributed on both the sides.This implies that the charge developed on each side will be

${q}_{1}={q}_{2}=\frac{Q}{2}$

This implies that the net surface charge density on each side $=\frac{\mathrm{Q}}{2\mathrm{A}}$

(b)

Electric field to the left of the plates

On the left side of the plate surface, charge density,

$\sigma =\frac{\mathrm{Q}}{2\mathrm{A}}$

Hence, electric field $=\frac{\mathrm{Q}}{2\mathrm{A}{\in}_{0}}$

This must be directed towards the left, as 'X' is the positively-charged plate.

(c) Here, the charged plate 'X' acts as the only source of electric field, with positive in the inner side. Plate Y is neutral. So, a negative charge will be induced on its inner side. 'Y' attracts the charged particle towards itself. So, the middle portion E is towards the right and is equal to $\frac{Q}{2A{\in}_{0}}$.

(d) Similarly for the extreme right, the outer side of plate 'Y' acts as positive and hence it repels to the right with $E\mathit{=}\frac{\mathit{Q}}{\mathit{2}\mathit{A}\mathit{}{\mathit{\in}}_{\mathit{0}}}$

#### Page No 142:

#### Answer:

Consider the Gaussian surface as shown in the figure.

Let the charge on the outer surface of the left-most plate be *q*. Thus, the charges on the plates are distributed as shown in the diagram.

The net field at point P due to all the induced charges must be zero, as it is lying inside the metal surface.

Let the surface area of the plates be *A*.

Electric field at point P due to the charges on plate X:

Due to charge (+*Q* $-$ *q*) is $\frac{Q-q}{2A{\in}_{0}}$ in the right direction

Due to charge (+*q*) is $\frac{q}{2A{\in}_{0}}$ in the right direction

Electric field at point P due to charges on plate Y:

Due to charge ($-$*q*) is $\frac{q}{2A{\in}_{0}}$ in the right direction

Due to charge (+*q*) is $\frac{q}{2A{\in}_{0}}$ in the left direction

Electric field at point P due to charges on plate Z:

Due to charge ($-$*q*) is $\frac{q}{2A{\in}_{0}}$ in the right direction

Due to charge ($-$2*Q** + q*) is $\frac{2Q-q}{2A{\in}_{0}}$ in the right direction

The net electric field at point P:

$\frac{Q-q}{2A{\in}_{0}}$ + $\frac{q}{2A{\in}_{0}}$ $-$ $\frac{q}{2A{\in}_{0}}$ $-$ $\frac{q}{2A{\in}_{0}}$ + $\frac{q}{2A{\in}_{0}}$ + $\frac{2Q-q}{2A{\in}_{0}}$ = 0

$\frac{Q-q}{2A{\in}_{0}}$ + $\frac{2Q-q}{2A{\in}_{0}}$ = 0

$Q-q+2Q-q=0\phantom{\rule{0ex}{0ex}}3Q-q=0\phantom{\rule{0ex}{0ex}}q=\frac{3Q}{2}$

Thus, the charge on the outer plate of the right-most plate = $-2Q+q=-2Q+\frac{3Q}{2}=-\frac{Q}{2}$

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