QBManager

PHXII01:ELECTRIC CHARGES AND FIELDS

357949 The adjacent diagram shows a charge + $Q$ held on an insulating support $S$ and enclosed by a hollow spherical conductor. $O$ represents the centre of the spherical conductor and $P$ is a point such that $O P = x a n d S P = r$ . The electric field at a point $P$ will be
supporting img

1

\frac{Q}{4 π ε_{0} r^{2}}

2

\frac{Q}{4 π ε_{0} x^{2}}

3 Zero

4 None of these

Explanation:

The charge present at $S$ induces charges -Q and +Q on the inner and outer surfaces of the spherical shell. The charge on the inner surface will be distributed unsymmetrically and on the outer surface symmetrically (due to shielding effect)
Due to the shielding effect by the conducting medium the effect of point charge $Q$ and induced charge -Q is zero outside the shell and also at point $P$ . The field at point $P$ is only because of +Q on the outer surface
$E = \frac{Q}{4 π ε_{0} x^{2}}$
supporting img

PHXII01:ELECTRIC CHARGES AND FIELDS

357950 A ball having a charge $- 100 e$ is placed at the centre of a metallic hollow spherical shell which has a net charge of $- 20 e$ . What is the charge on the shell's outer surface?

1

+ 80 e

2

- 20 e

3

- 100 e

4

- 120 e

Explanation:

supporting img
When charges ball is placed at the center of a ball it induces opposite charge on the inner surface by the induction further same nature charge will induce on the outer surface of shell. Hence the total charge on the outer surface of shell is $- 100 e - 20 e = - 120 e$

PHXII01:ELECTRIC CHARGES AND FIELDS

357951 An early model for an atom considered it to have a positively charged point nucleus of charge $Z e$ , surrounded by a uniform density of negative charge upto a radius $R$ . The atom as a whole is neutral. The electric field at a distance $r$ from the nucleus is $(r < R)$
supporting img

1

\frac{Z e}{4 π ε_{0}} [\frac{1}{r^{2}} - \frac{r}{R^{2}}]

2

\frac{Z e}{4 π ε_{0}} [\frac{1}{r^{2}} - \frac{r}{R^{3}}]

3

\frac{Z e}{4 π ε_{0}} [\frac{r}{R^{3}} + \frac{1}{r^{2}}]

4

\frac{Z e}{4 π ε_{0}} [\frac{r}{R^{3}} - \frac{1}{r^{2}}]

Explanation:

Charge on nucleus $= + Z e$
Total negative charge $= - Z e$
Negative charge density,
$ρ = \frac{C h a r g e}{V o l u m e} = \frac{- Z e}{\frac{4}{3} π R^{3}}$
i.e., $ρ = - \frac{3}{4} \frac{Z e}{π R^{3}} (1)$
Consider a Gaussian surface with radius $r$ .
By Gauss’s theorem
$ϕ = E (r) \times 4 π r^{2} = \frac{q}{ε_{0}} (2)$
Charge enclosed by Gaussian surface
$q = Z e + \frac{4 π r^{3}}{3} ρ = Z e - Z e \frac{r^{3}}{R^{3}}$
$E (r) = \frac{q}{4 π ε_{0} r^{2}} = \frac{Z e - Z e \frac{r^{3}}{R^{3}}}{4 π ε_{0} r^{2}} = \frac{Z e}{4 π ε_{0}} [\frac{1}{r^{2}} - \frac{r}{R^{3}}]$

PHXII01:ELECTRIC CHARGES AND FIELDS

357952 $S_{1}$ and $S_{2}$ are two hollow concentric spheres with charge $q$ and $2 q$ . Space between $S_{1}$ and $S_{2}$ is filled with a dielectric of dielectric constants.
The ratio of flux through $S_{2}$ and flux through $S_{1}$ is $K$ . Then, find the value of $\frac{9 K}{11}$
supporting img

1 9

2 15

3 5

4 10

Explanation:

With air in the space, between $S_{1}$ and $S_{2}$ .
$ϕ_{S_{1}} = \frac{q}{ε_{0}} and ϕ_{S_{2}} = \frac{2 q}{ε_{0}}$
When a medium of $K = 5$ is filled in space
$ϕ_{S_{1}}^{'} = \frac{Q}{K ε_{0}} = \frac{ϕ_{S_{1}}}{5} = \frac{q}{5 ε_{0}}$
And in this case, flux though $S_{2}$
$ϕ_{S_{2}}^{'} = flux of S_{1} + flux of charge 2 q$
$= \frac{q}{5 ε_{0}} + \frac{2 q}{ε_{0}} = \frac{q + 10 q}{5 ε_{0}} = \frac{11 q}{5 ε_{0}}$
So, ratio $\frac{ϕ_{S_{1}}^{'}}{ϕ_{S_{2}}^{'}} = 11$
So, $\frac{9 K}{11} = 9$

PHXII01:ELECTRIC CHARGES AND FIELDS

357953 A sphere $S_{1}$ of radius $R$ , encloses a total charge $q$ . If there is another concentric sphere $S_{2}$ of radius $2 R$ and there be no additional charges between $S_{1}$ and $S_{2}$ . What is the ratio of electric flux through $S_{1}$ and $S_{2}$ ?

1 1.5

2 2

3 1

4 0.5

Explanation:

$ϕ = \frac{q}{ε_{0}}$
So,
$\frac{ϕ_{1}}{ϕ_{2}} = \frac{q_{1}}{ε_{0}} \times \frac{ε_{0}}{q_{2}} = \frac{q_{1}}{q_{2}}$
$\Rightarrow \frac{ϕ_{1}}{ϕ_{2}} = \frac{q}{q} = 1$ .

PHXII01:ELECTRIC CHARGES AND FIELDS

357949 The adjacent diagram shows a charge + $Q$ held on an insulating support $S$ and enclosed by a hollow spherical conductor. $O$ represents the centre of the spherical conductor and $P$ is a point such that $O P = x a n d S P = r$ . The electric field at a point $P$ will be
supporting img

1

\frac{Q}{4 π ε_{0} r^{2}}

2

\frac{Q}{4 π ε_{0} x^{2}}

3 Zero

4 None of these

Explanation:

The charge present at $S$ induces charges -Q and +Q on the inner and outer surfaces of the spherical shell. The charge on the inner surface will be distributed unsymmetrically and on the outer surface symmetrically (due to shielding effect)
Due to the shielding effect by the conducting medium the effect of point charge $Q$ and induced charge -Q is zero outside the shell and also at point $P$ . The field at point $P$ is only because of +Q on the outer surface
$E = \frac{Q}{4 π ε_{0} x^{2}}$
supporting img

PHXII01:ELECTRIC CHARGES AND FIELDS

357950 A ball having a charge $- 100 e$ is placed at the centre of a metallic hollow spherical shell which has a net charge of $- 20 e$ . What is the charge on the shell's outer surface?

1

+ 80 e

2

- 20 e

3

- 100 e

4

- 120 e

Explanation:

supporting img
When charges ball is placed at the center of a ball it induces opposite charge on the inner surface by the induction further same nature charge will induce on the outer surface of shell. Hence the total charge on the outer surface of shell is $- 100 e - 20 e = - 120 e$

PHXII01:ELECTRIC CHARGES AND FIELDS

357951 An early model for an atom considered it to have a positively charged point nucleus of charge $Z e$ , surrounded by a uniform density of negative charge upto a radius $R$ . The atom as a whole is neutral. The electric field at a distance $r$ from the nucleus is $(r < R)$
supporting img

1

\frac{Z e}{4 π ε_{0}} [\frac{1}{r^{2}} - \frac{r}{R^{2}}]

2

\frac{Z e}{4 π ε_{0}} [\frac{1}{r^{2}} - \frac{r}{R^{3}}]

3

\frac{Z e}{4 π ε_{0}} [\frac{r}{R^{3}} + \frac{1}{r^{2}}]

4

\frac{Z e}{4 π ε_{0}} [\frac{r}{R^{3}} - \frac{1}{r^{2}}]

Explanation:

Charge on nucleus $= + Z e$
Total negative charge $= - Z e$
Negative charge density,
$ρ = \frac{C h a r g e}{V o l u m e} = \frac{- Z e}{\frac{4}{3} π R^{3}}$
i.e., $ρ = - \frac{3}{4} \frac{Z e}{π R^{3}} (1)$
Consider a Gaussian surface with radius $r$ .
By Gauss’s theorem
$ϕ = E (r) \times 4 π r^{2} = \frac{q}{ε_{0}} (2)$
Charge enclosed by Gaussian surface
$q = Z e + \frac{4 π r^{3}}{3} ρ = Z e - Z e \frac{r^{3}}{R^{3}}$
$E (r) = \frac{q}{4 π ε_{0} r^{2}} = \frac{Z e - Z e \frac{r^{3}}{R^{3}}}{4 π ε_{0} r^{2}} = \frac{Z e}{4 π ε_{0}} [\frac{1}{r^{2}} - \frac{r}{R^{3}}]$

PHXII01:ELECTRIC CHARGES AND FIELDS

357952 $S_{1}$ and $S_{2}$ are two hollow concentric spheres with charge $q$ and $2 q$ . Space between $S_{1}$ and $S_{2}$ is filled with a dielectric of dielectric constants.
The ratio of flux through $S_{2}$ and flux through $S_{1}$ is $K$ . Then, find the value of $\frac{9 K}{11}$
supporting img

1 9

2 15

3 5

4 10

Explanation:

With air in the space, between $S_{1}$ and $S_{2}$ .
$ϕ_{S_{1}} = \frac{q}{ε_{0}} and ϕ_{S_{2}} = \frac{2 q}{ε_{0}}$
When a medium of $K = 5$ is filled in space
$ϕ_{S_{1}}^{'} = \frac{Q}{K ε_{0}} = \frac{ϕ_{S_{1}}}{5} = \frac{q}{5 ε_{0}}$
And in this case, flux though $S_{2}$
$ϕ_{S_{2}}^{'} = flux of S_{1} + flux of charge 2 q$
$= \frac{q}{5 ε_{0}} + \frac{2 q}{ε_{0}} = \frac{q + 10 q}{5 ε_{0}} = \frac{11 q}{5 ε_{0}}$
So, ratio $\frac{ϕ_{S_{1}}^{'}}{ϕ_{S_{2}}^{'}} = 11$
So, $\frac{9 K}{11} = 9$

PHXII01:ELECTRIC CHARGES AND FIELDS

357953 A sphere $S_{1}$ of radius $R$ , encloses a total charge $q$ . If there is another concentric sphere $S_{2}$ of radius $2 R$ and there be no additional charges between $S_{1}$ and $S_{2}$ . What is the ratio of electric flux through $S_{1}$ and $S_{2}$ ?

1 1.5

2 2

3 1

4 0.5

Explanation:

$ϕ = \frac{q}{ε_{0}}$
So,
$\frac{ϕ_{1}}{ϕ_{2}} = \frac{q_{1}}{ε_{0}} \times \frac{ε_{0}}{q_{2}} = \frac{q_{1}}{q_{2}}$
$\Rightarrow \frac{ϕ_{1}}{ϕ_{2}} = \frac{q}{q} = 1$ .

PHXII01:ELECTRIC CHARGES AND FIELDS

357949 The adjacent diagram shows a charge + $Q$ held on an insulating support $S$ and enclosed by a hollow spherical conductor. $O$ represents the centre of the spherical conductor and $P$ is a point such that $O P = x a n d S P = r$ . The electric field at a point $P$ will be
supporting img

1

\frac{Q}{4 π ε_{0} r^{2}}

2

\frac{Q}{4 π ε_{0} x^{2}}

3 Zero

4 None of these

Explanation:

The charge present at $S$ induces charges -Q and +Q on the inner and outer surfaces of the spherical shell. The charge on the inner surface will be distributed unsymmetrically and on the outer surface symmetrically (due to shielding effect)
Due to the shielding effect by the conducting medium the effect of point charge $Q$ and induced charge -Q is zero outside the shell and also at point $P$ . The field at point $P$ is only because of +Q on the outer surface
$E = \frac{Q}{4 π ε_{0} x^{2}}$
supporting img

PHXII01:ELECTRIC CHARGES AND FIELDS

357950 A ball having a charge $- 100 e$ is placed at the centre of a metallic hollow spherical shell which has a net charge of $- 20 e$ . What is the charge on the shell's outer surface?

1

+ 80 e

2

- 20 e

3

- 100 e

4

- 120 e

Explanation:

supporting img
When charges ball is placed at the center of a ball it induces opposite charge on the inner surface by the induction further same nature charge will induce on the outer surface of shell. Hence the total charge on the outer surface of shell is $- 100 e - 20 e = - 120 e$

PHXII01:ELECTRIC CHARGES AND FIELDS

357951 An early model for an atom considered it to have a positively charged point nucleus of charge $Z e$ , surrounded by a uniform density of negative charge upto a radius $R$ . The atom as a whole is neutral. The electric field at a distance $r$ from the nucleus is $(r < R)$
supporting img

1

\frac{Z e}{4 π ε_{0}} [\frac{1}{r^{2}} - \frac{r}{R^{2}}]

2

\frac{Z e}{4 π ε_{0}} [\frac{1}{r^{2}} - \frac{r}{R^{3}}]

3

\frac{Z e}{4 π ε_{0}} [\frac{r}{R^{3}} + \frac{1}{r^{2}}]

4

\frac{Z e}{4 π ε_{0}} [\frac{r}{R^{3}} - \frac{1}{r^{2}}]

Explanation:

Charge on nucleus $= + Z e$
Total negative charge $= - Z e$
Negative charge density,
$ρ = \frac{C h a r g e}{V o l u m e} = \frac{- Z e}{\frac{4}{3} π R^{3}}$
i.e., $ρ = - \frac{3}{4} \frac{Z e}{π R^{3}} (1)$
Consider a Gaussian surface with radius $r$ .
By Gauss’s theorem
$ϕ = E (r) \times 4 π r^{2} = \frac{q}{ε_{0}} (2)$
Charge enclosed by Gaussian surface
$q = Z e + \frac{4 π r^{3}}{3} ρ = Z e - Z e \frac{r^{3}}{R^{3}}$
$E (r) = \frac{q}{4 π ε_{0} r^{2}} = \frac{Z e - Z e \frac{r^{3}}{R^{3}}}{4 π ε_{0} r^{2}} = \frac{Z e}{4 π ε_{0}} [\frac{1}{r^{2}} - \frac{r}{R^{3}}]$

PHXII01:ELECTRIC CHARGES AND FIELDS

357952 $S_{1}$ and $S_{2}$ are two hollow concentric spheres with charge $q$ and $2 q$ . Space between $S_{1}$ and $S_{2}$ is filled with a dielectric of dielectric constants.
The ratio of flux through $S_{2}$ and flux through $S_{1}$ is $K$ . Then, find the value of $\frac{9 K}{11}$
supporting img

1 9

2 15

3 5

4 10

Explanation:

With air in the space, between $S_{1}$ and $S_{2}$ .
$ϕ_{S_{1}} = \frac{q}{ε_{0}} and ϕ_{S_{2}} = \frac{2 q}{ε_{0}}$
When a medium of $K = 5$ is filled in space
$ϕ_{S_{1}}^{'} = \frac{Q}{K ε_{0}} = \frac{ϕ_{S_{1}}}{5} = \frac{q}{5 ε_{0}}$
And in this case, flux though $S_{2}$
$ϕ_{S_{2}}^{'} = flux of S_{1} + flux of charge 2 q$
$= \frac{q}{5 ε_{0}} + \frac{2 q}{ε_{0}} = \frac{q + 10 q}{5 ε_{0}} = \frac{11 q}{5 ε_{0}}$
So, ratio $\frac{ϕ_{S_{1}}^{'}}{ϕ_{S_{2}}^{'}} = 11$
So, $\frac{9 K}{11} = 9$

PHXII01:ELECTRIC CHARGES AND FIELDS

357953 A sphere $S_{1}$ of radius $R$ , encloses a total charge $q$ . If there is another concentric sphere $S_{2}$ of radius $2 R$ and there be no additional charges between $S_{1}$ and $S_{2}$ . What is the ratio of electric flux through $S_{1}$ and $S_{2}$ ?

1 1.5

2 2

3 1

4 0.5

Explanation:

$ϕ = \frac{q}{ε_{0}}$
So,
$\frac{ϕ_{1}}{ϕ_{2}} = \frac{q_{1}}{ε_{0}} \times \frac{ε_{0}}{q_{2}} = \frac{q_{1}}{q_{2}}$
$\Rightarrow \frac{ϕ_{1}}{ϕ_{2}} = \frac{q}{q} = 1$ .

PHXII01:ELECTRIC CHARGES AND FIELDS

357949 The adjacent diagram shows a charge + $Q$ held on an insulating support $S$ and enclosed by a hollow spherical conductor. $O$ represents the centre of the spherical conductor and $P$ is a point such that $O P = x a n d S P = r$ . The electric field at a point $P$ will be
supporting img

1

\frac{Q}{4 π ε_{0} r^{2}}

2

\frac{Q}{4 π ε_{0} x^{2}}

3 Zero

4 None of these

Explanation:

The charge present at $S$ induces charges -Q and +Q on the inner and outer surfaces of the spherical shell. The charge on the inner surface will be distributed unsymmetrically and on the outer surface symmetrically (due to shielding effect)
Due to the shielding effect by the conducting medium the effect of point charge $Q$ and induced charge -Q is zero outside the shell and also at point $P$ . The field at point $P$ is only because of +Q on the outer surface
$E = \frac{Q}{4 π ε_{0} x^{2}}$
supporting img

PHXII01:ELECTRIC CHARGES AND FIELDS

357950 A ball having a charge $- 100 e$ is placed at the centre of a metallic hollow spherical shell which has a net charge of $- 20 e$ . What is the charge on the shell's outer surface?

1

+ 80 e

2

- 20 e

3

- 100 e

4

- 120 e

Explanation:

supporting img
When charges ball is placed at the center of a ball it induces opposite charge on the inner surface by the induction further same nature charge will induce on the outer surface of shell. Hence the total charge on the outer surface of shell is $- 100 e - 20 e = - 120 e$

PHXII01:ELECTRIC CHARGES AND FIELDS

357951 An early model for an atom considered it to have a positively charged point nucleus of charge $Z e$ , surrounded by a uniform density of negative charge upto a radius $R$ . The atom as a whole is neutral. The electric field at a distance $r$ from the nucleus is $(r < R)$
supporting img

1

\frac{Z e}{4 π ε_{0}} [\frac{1}{r^{2}} - \frac{r}{R^{2}}]

2

\frac{Z e}{4 π ε_{0}} [\frac{1}{r^{2}} - \frac{r}{R^{3}}]

3

\frac{Z e}{4 π ε_{0}} [\frac{r}{R^{3}} + \frac{1}{r^{2}}]

4

\frac{Z e}{4 π ε_{0}} [\frac{r}{R^{3}} - \frac{1}{r^{2}}]

Explanation:

Charge on nucleus $= + Z e$
Total negative charge $= - Z e$
Negative charge density,
$ρ = \frac{C h a r g e}{V o l u m e} = \frac{- Z e}{\frac{4}{3} π R^{3}}$
i.e., $ρ = - \frac{3}{4} \frac{Z e}{π R^{3}} (1)$
Consider a Gaussian surface with radius $r$ .
By Gauss’s theorem
$ϕ = E (r) \times 4 π r^{2} = \frac{q}{ε_{0}} (2)$
Charge enclosed by Gaussian surface
$q = Z e + \frac{4 π r^{3}}{3} ρ = Z e - Z e \frac{r^{3}}{R^{3}}$
$E (r) = \frac{q}{4 π ε_{0} r^{2}} = \frac{Z e - Z e \frac{r^{3}}{R^{3}}}{4 π ε_{0} r^{2}} = \frac{Z e}{4 π ε_{0}} [\frac{1}{r^{2}} - \frac{r}{R^{3}}]$

PHXII01:ELECTRIC CHARGES AND FIELDS

357952 $S_{1}$ and $S_{2}$ are two hollow concentric spheres with charge $q$ and $2 q$ . Space between $S_{1}$ and $S_{2}$ is filled with a dielectric of dielectric constants.
The ratio of flux through $S_{2}$ and flux through $S_{1}$ is $K$ . Then, find the value of $\frac{9 K}{11}$
supporting img

1 9

2 15

3 5

4 10

Explanation:

With air in the space, between $S_{1}$ and $S_{2}$ .
$ϕ_{S_{1}} = \frac{q}{ε_{0}} and ϕ_{S_{2}} = \frac{2 q}{ε_{0}}$
When a medium of $K = 5$ is filled in space
$ϕ_{S_{1}}^{'} = \frac{Q}{K ε_{0}} = \frac{ϕ_{S_{1}}}{5} = \frac{q}{5 ε_{0}}$
And in this case, flux though $S_{2}$
$ϕ_{S_{2}}^{'} = flux of S_{1} + flux of charge 2 q$
$= \frac{q}{5 ε_{0}} + \frac{2 q}{ε_{0}} = \frac{q + 10 q}{5 ε_{0}} = \frac{11 q}{5 ε_{0}}$
So, ratio $\frac{ϕ_{S_{1}}^{'}}{ϕ_{S_{2}}^{'}} = 11$
So, $\frac{9 K}{11} = 9$

PHXII01:ELECTRIC CHARGES AND FIELDS

357953 A sphere $S_{1}$ of radius $R$ , encloses a total charge $q$ . If there is another concentric sphere $S_{2}$ of radius $2 R$ and there be no additional charges between $S_{1}$ and $S_{2}$ . What is the ratio of electric flux through $S_{1}$ and $S_{2}$ ?

1 1.5

2 2

3 1

4 0.5

Explanation:

$ϕ = \frac{q}{ε_{0}}$
So,
$\frac{ϕ_{1}}{ϕ_{2}} = \frac{q_{1}}{ε_{0}} \times \frac{ε_{0}}{q_{2}} = \frac{q_{1}}{q_{2}}$
$\Rightarrow \frac{ϕ_{1}}{ϕ_{2}} = \frac{q}{q} = 1$ .

PHXII01:ELECTRIC CHARGES AND FIELDS

357949 The adjacent diagram shows a charge + $Q$ held on an insulating support $S$ and enclosed by a hollow spherical conductor. $O$ represents the centre of the spherical conductor and $P$ is a point such that $O P = x a n d S P = r$ . The electric field at a point $P$ will be
supporting img

1

\frac{Q}{4 π ε_{0} r^{2}}

2

\frac{Q}{4 π ε_{0} x^{2}}

3 Zero

4 None of these

Explanation:

The charge present at $S$ induces charges -Q and +Q on the inner and outer surfaces of the spherical shell. The charge on the inner surface will be distributed unsymmetrically and on the outer surface symmetrically (due to shielding effect)
Due to the shielding effect by the conducting medium the effect of point charge $Q$ and induced charge -Q is zero outside the shell and also at point $P$ . The field at point $P$ is only because of +Q on the outer surface
$E = \frac{Q}{4 π ε_{0} x^{2}}$
supporting img

PHXII01:ELECTRIC CHARGES AND FIELDS

357950 A ball having a charge $- 100 e$ is placed at the centre of a metallic hollow spherical shell which has a net charge of $- 20 e$ . What is the charge on the shell's outer surface?

1

+ 80 e

2

- 20 e

3

- 100 e

4

- 120 e

Explanation:

supporting img
When charges ball is placed at the center of a ball it induces opposite charge on the inner surface by the induction further same nature charge will induce on the outer surface of shell. Hence the total charge on the outer surface of shell is $- 100 e - 20 e = - 120 e$

PHXII01:ELECTRIC CHARGES AND FIELDS

357951 An early model for an atom considered it to have a positively charged point nucleus of charge $Z e$ , surrounded by a uniform density of negative charge upto a radius $R$ . The atom as a whole is neutral. The electric field at a distance $r$ from the nucleus is $(r < R)$
supporting img

1

\frac{Z e}{4 π ε_{0}} [\frac{1}{r^{2}} - \frac{r}{R^{2}}]

2

\frac{Z e}{4 π ε_{0}} [\frac{1}{r^{2}} - \frac{r}{R^{3}}]

3

\frac{Z e}{4 π ε_{0}} [\frac{r}{R^{3}} + \frac{1}{r^{2}}]

4

\frac{Z e}{4 π ε_{0}} [\frac{r}{R^{3}} - \frac{1}{r^{2}}]

Explanation:

Charge on nucleus $= + Z e$
Total negative charge $= - Z e$
Negative charge density,
$ρ = \frac{C h a r g e}{V o l u m e} = \frac{- Z e}{\frac{4}{3} π R^{3}}$
i.e., $ρ = - \frac{3}{4} \frac{Z e}{π R^{3}} (1)$
Consider a Gaussian surface with radius $r$ .
By Gauss’s theorem
$ϕ = E (r) \times 4 π r^{2} = \frac{q}{ε_{0}} (2)$
Charge enclosed by Gaussian surface
$q = Z e + \frac{4 π r^{3}}{3} ρ = Z e - Z e \frac{r^{3}}{R^{3}}$
$E (r) = \frac{q}{4 π ε_{0} r^{2}} = \frac{Z e - Z e \frac{r^{3}}{R^{3}}}{4 π ε_{0} r^{2}} = \frac{Z e}{4 π ε_{0}} [\frac{1}{r^{2}} - \frac{r}{R^{3}}]$

PHXII01:ELECTRIC CHARGES AND FIELDS

357952 $S_{1}$ and $S_{2}$ are two hollow concentric spheres with charge $q$ and $2 q$ . Space between $S_{1}$ and $S_{2}$ is filled with a dielectric of dielectric constants.
The ratio of flux through $S_{2}$ and flux through $S_{1}$ is $K$ . Then, find the value of $\frac{9 K}{11}$
supporting img

1 9

2 15

3 5

4 10

Explanation:

With air in the space, between $S_{1}$ and $S_{2}$ .
$ϕ_{S_{1}} = \frac{q}{ε_{0}} and ϕ_{S_{2}} = \frac{2 q}{ε_{0}}$
When a medium of $K = 5$ is filled in space
$ϕ_{S_{1}}^{'} = \frac{Q}{K ε_{0}} = \frac{ϕ_{S_{1}}}{5} = \frac{q}{5 ε_{0}}$
And in this case, flux though $S_{2}$
$ϕ_{S_{2}}^{'} = flux of S_{1} + flux of charge 2 q$
$= \frac{q}{5 ε_{0}} + \frac{2 q}{ε_{0}} = \frac{q + 10 q}{5 ε_{0}} = \frac{11 q}{5 ε_{0}}$
So, ratio $\frac{ϕ_{S_{1}}^{'}}{ϕ_{S_{2}}^{'}} = 11$
So, $\frac{9 K}{11} = 9$

PHXII01:ELECTRIC CHARGES AND FIELDS

357953 A sphere $S_{1}$ of radius $R$ , encloses a total charge $q$ . If there is another concentric sphere $S_{2}$ of radius $2 R$ and there be no additional charges between $S_{1}$ and $S_{2}$ . What is the ratio of electric flux through $S_{1}$ and $S_{2}$ ?

1 1.5

2 2

3 1

4 0.5

Explanation:

$ϕ = \frac{q}{ε_{0}}$
So,
$\frac{ϕ_{1}}{ϕ_{2}} = \frac{q_{1}}{ε_{0}} \times \frac{ε_{0}}{q_{2}} = \frac{q_{1}}{q_{2}}$
$\Rightarrow \frac{ϕ_{1}}{ϕ_{2}} = \frac{q}{q} = 1$ .

Explanation:

Explanation:

Explanation:

Explanation:

Explanation:

Explanation:

Explanation:

Explanation:

Explanation:

Explanation:

Explanation:

Explanation:

Explanation:

Explanation:

Explanation:

Question Bank Series

Explanation:

Explanation:

Explanation:

Explanation:

Explanation:

Explanation:

Explanation:

Explanation:

Explanation:

Explanation: