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PHXII01:ELECTRIC CHARGES AND FIELDS

358015 A positively charged sphere of radius $r_{0}$ carries a volume charge density $ρ$ . A spherical cavity of radius $r_{0} / 2$ is then scooped out and left empty. $C_{1}$ is the centre of the sphere and $C_{2}$ that of the cavity. What is the direction and magnitude of the electric field at point $B$ ?
supporting img

1

\frac{17 ρ r_{0}}{54 ε_{0}} l e f t

2

\frac{ρ r_{0}}{6 ε_{0}} l e f t

3

\frac{17 ρ r_{0}}{54 ε_{0}} r i g h t

4

\frac{ρ r_{0}}{6 ε_{0}} r i g h t

Explanation:

The electric field at point $B$ is given by
${\vec{E}}_{o r i g i n a l} = {\vec{E}}_{r e m o v e d} + {\vec{E}}_{r e m a i n i n g}$
${\vec{E}}_{o r i g i n a l}$ – Electric field by the original sphere without cavity
${\vec{E}}_{r e m o v e d}$ – Electric field by the removed part
${\vec{E}}_{r e m a i n i n g}$ – Electric field by the remaining part.
${\vec{E}}_{o r i g i n a l} = \frac{- ρ (\frac{4}{3} π r_{0}^{3})}{4 π ε_{0} r_{0}^{2}} \hat{i} = \frac{- ρ r_{0}}{3 ε_{0}} \hat{i}$
${\vec{E}}_{r e m o v e d} = \frac{- ρ \frac{4}{3} π {(\frac{r_{0}}{2})}^{3}}{4 π ε_{0} {(\frac{3 r_{0}}{2})}^{2}} = \frac{- ρ r_{0}}{54 ε_{0}} \hat{i}$
${\vec{E}}_{r e m a i n i n g} = \frac{- 17 ρ r_{0}}{54 ε_{0}} \hat{i}$ (Left to the point $B$ )

PHXII01:ELECTRIC CHARGES AND FIELDS

358016 A conducting sphere of radius $20 c m$ has a known charge. If the electric field at a distance $40 c m$ from the centre of the sphere is $1.2 \times 10^{3} N C^{- 1}$ and points radially inwards. The net charge on the sphere is:

1

- 4.5 \times 10^{- 9} C

2

4.5 \times 10^{9} C

3

- 21.3 \times 10^{- 9} C

4

21.3 \times 10^{9} C

Explanation:

supporting img
$r = 40 cm, \vec{E} = 1.2 \times 10^{3} N / C, \vec{E} = \frac{- K q}{r^{2}}$
$q = \frac{- 1.2 \times 10^{3} \times \frac{40}{100} \times \frac{40}{100}}{9 \times 10^{9}} = - 21.3 \times 10^{- 19} C$

PHXII01:ELECTRIC CHARGES AND FIELDS

358017 A spherical portion has been removed from a solid sphere having a charge distributed uniformly in its volume in the figure. The electric field inside the emptied space is
supporting img

1 Zero everywhere

2 Uniform

3 Non-uniform

4 Zero only at its center

Explanation:

Let $O_{1} a n d O_{2}$ are centres of the sphere and the cavity. Let $P$ is an arbitary point in the spherical cavity.
The electric field inside a solid sphere of total charge $Q$ and radius $R$ at a radial $r$ from the centre is
$E = \frac{Q r}{4 π ε_{0} R^{3}} = \frac{ρ r}{3 ε_{0}} (ρ = \frac{Q}{\frac{4}{3} π R^{3}})$
In vector form $\overset{―}{E} = \frac{ρ \overset{―}{r}}{3 ε_{0}}$
The electric field at $P$ is given as
${\overset{―}{E}}_{o r i g i n a l} = {\overset{―}{E}}_{r e m o v e d} + {\overset{―}{E}}_{r e m a i n i n g} = {\overset{―}{E}}_{r e m o v e d} + {\overset{―}{E}}_{p}$
Where ${\overset{―}{E}}_{P}$ - Electric field at point $P$ because of the remaining part.
${\overset{―}{E}}_{o r i g i n a l} = \frac{ρ \overset{―}{O_{1} P}}{3 ε_{0}}, {\overset{―}{E}}_{r e m o v e d} = \frac{ρ \overset{―}{O_{2} P}}{3 ε_{0}}$
$= \frac{ρ}{3 ε_{0}} (\overset{―}{O_{1} P} - \overset{―}{O_{2} P}) = \frac{ρ \overset{―}{O_{2} O_{2}}}{3 ε_{0}}$
The electric field inside the cavity is uniform and its direction is present along $O_{1} O_{2}$
supporting img

PHXII01:ELECTRIC CHARGES AND FIELDS

358018 $4 \times 10^{10}$ electrons are removed from a neutral metal sphere of diameter $20 c m$ placed in air. The magnitude of the electric field (in $N C^{- 1}$ ) at a distance of 20 $c m$ from its centre is

1

z e r o

2

5760

3

640

4

1440

Explanation:

Number of electrons removed $= 4 \times 10^{10}$
Charge on this sphere $= 4 \times 10^{10} \times 1.6 \times 10^{- 19} C$
$= 6.4 \times 10^{- 19} C = 6.4 n C$
Diameter of Sphere, $D = 20 c m = 0.2 m$
Distance from the centre, $d = 20 c m = 0.2 m$
Electric field at point p, $E = \frac{q}{4 π ε_{0} d^{2}}$
$(\frac{9 \times 10^{9} \times 6.4 \times 10^{- 9}}{{(0.2)}^{2}}) N C^{- 1} = 1440 N C^{- 1}$
supporting img

PHXII01:ELECTRIC CHARGES AND FIELDS

358019 $σ$ is the uniform surface charge density of a thin spherical shell of radius $R$ . The electric field at any point on the surface of the spherical shell is

1

σ / ε_{0}

2

σ / 4 ε_{0}

3

σ / ε_{0} R

4

σ / 2 ε_{0}

Explanation:

Electric field at any point on the surface of a thin spherical shell is, $E = \frac{1}{4 π ε_{0}} \cdot \frac{Q}{R^{2}} (1)$
Now, charge on the spherical shell, is,
$Q =$ surface charge density $\times$ Area of spherical shell
supporting img

$∴ Q = σ \times 4 π R^{2}$
Substituting the value of $Q$ in equation (1), we
get $E = \frac{1}{4 π ε_{0}} \times \frac{σ \times 4 π R^{2}}{R^{2}} = \frac{σ}{ε_{0}}$

JEE - 2024

PHXII01:ELECTRIC CHARGES AND FIELDS

358015 A positively charged sphere of radius $r_{0}$ carries a volume charge density $ρ$ . A spherical cavity of radius $r_{0} / 2$ is then scooped out and left empty. $C_{1}$ is the centre of the sphere and $C_{2}$ that of the cavity. What is the direction and magnitude of the electric field at point $B$ ?
supporting img

1

\frac{17 ρ r_{0}}{54 ε_{0}} l e f t

2

\frac{ρ r_{0}}{6 ε_{0}} l e f t

3

\frac{17 ρ r_{0}}{54 ε_{0}} r i g h t

4

\frac{ρ r_{0}}{6 ε_{0}} r i g h t

Explanation:

The electric field at point $B$ is given by
${\vec{E}}_{o r i g i n a l} = {\vec{E}}_{r e m o v e d} + {\vec{E}}_{r e m a i n i n g}$
${\vec{E}}_{o r i g i n a l}$ – Electric field by the original sphere without cavity
${\vec{E}}_{r e m o v e d}$ – Electric field by the removed part
${\vec{E}}_{r e m a i n i n g}$ – Electric field by the remaining part.
${\vec{E}}_{o r i g i n a l} = \frac{- ρ (\frac{4}{3} π r_{0}^{3})}{4 π ε_{0} r_{0}^{2}} \hat{i} = \frac{- ρ r_{0}}{3 ε_{0}} \hat{i}$
${\vec{E}}_{r e m o v e d} = \frac{- ρ \frac{4}{3} π {(\frac{r_{0}}{2})}^{3}}{4 π ε_{0} {(\frac{3 r_{0}}{2})}^{2}} = \frac{- ρ r_{0}}{54 ε_{0}} \hat{i}$
${\vec{E}}_{r e m a i n i n g} = \frac{- 17 ρ r_{0}}{54 ε_{0}} \hat{i}$ (Left to the point $B$ )

PHXII01:ELECTRIC CHARGES AND FIELDS

358016 A conducting sphere of radius $20 c m$ has a known charge. If the electric field at a distance $40 c m$ from the centre of the sphere is $1.2 \times 10^{3} N C^{- 1}$ and points radially inwards. The net charge on the sphere is:

1

- 4.5 \times 10^{- 9} C

2

4.5 \times 10^{9} C

3

- 21.3 \times 10^{- 9} C

4

21.3 \times 10^{9} C

Explanation:

supporting img
$r = 40 cm, \vec{E} = 1.2 \times 10^{3} N / C, \vec{E} = \frac{- K q}{r^{2}}$
$q = \frac{- 1.2 \times 10^{3} \times \frac{40}{100} \times \frac{40}{100}}{9 \times 10^{9}} = - 21.3 \times 10^{- 19} C$

PHXII01:ELECTRIC CHARGES AND FIELDS

358017 A spherical portion has been removed from a solid sphere having a charge distributed uniformly in its volume in the figure. The electric field inside the emptied space is
supporting img

1 Zero everywhere

2 Uniform

3 Non-uniform

4 Zero only at its center

Explanation:

Let $O_{1} a n d O_{2}$ are centres of the sphere and the cavity. Let $P$ is an arbitary point in the spherical cavity.
The electric field inside a solid sphere of total charge $Q$ and radius $R$ at a radial $r$ from the centre is
$E = \frac{Q r}{4 π ε_{0} R^{3}} = \frac{ρ r}{3 ε_{0}} (ρ = \frac{Q}{\frac{4}{3} π R^{3}})$
In vector form $\overset{―}{E} = \frac{ρ \overset{―}{r}}{3 ε_{0}}$
The electric field at $P$ is given as
${\overset{―}{E}}_{o r i g i n a l} = {\overset{―}{E}}_{r e m o v e d} + {\overset{―}{E}}_{r e m a i n i n g} = {\overset{―}{E}}_{r e m o v e d} + {\overset{―}{E}}_{p}$
Where ${\overset{―}{E}}_{P}$ - Electric field at point $P$ because of the remaining part.
${\overset{―}{E}}_{o r i g i n a l} = \frac{ρ \overset{―}{O_{1} P}}{3 ε_{0}}, {\overset{―}{E}}_{r e m o v e d} = \frac{ρ \overset{―}{O_{2} P}}{3 ε_{0}}$
$= \frac{ρ}{3 ε_{0}} (\overset{―}{O_{1} P} - \overset{―}{O_{2} P}) = \frac{ρ \overset{―}{O_{2} O_{2}}}{3 ε_{0}}$
The electric field inside the cavity is uniform and its direction is present along $O_{1} O_{2}$
supporting img

PHXII01:ELECTRIC CHARGES AND FIELDS

358018 $4 \times 10^{10}$ electrons are removed from a neutral metal sphere of diameter $20 c m$ placed in air. The magnitude of the electric field (in $N C^{- 1}$ ) at a distance of 20 $c m$ from its centre is

1

z e r o

2

5760

3

640

4

1440

Explanation:

Number of electrons removed $= 4 \times 10^{10}$
Charge on this sphere $= 4 \times 10^{10} \times 1.6 \times 10^{- 19} C$
$= 6.4 \times 10^{- 19} C = 6.4 n C$
Diameter of Sphere, $D = 20 c m = 0.2 m$
Distance from the centre, $d = 20 c m = 0.2 m$
Electric field at point p, $E = \frac{q}{4 π ε_{0} d^{2}}$
$(\frac{9 \times 10^{9} \times 6.4 \times 10^{- 9}}{{(0.2)}^{2}}) N C^{- 1} = 1440 N C^{- 1}$
supporting img

PHXII01:ELECTRIC CHARGES AND FIELDS

358019 $σ$ is the uniform surface charge density of a thin spherical shell of radius $R$ . The electric field at any point on the surface of the spherical shell is

1

σ / ε_{0}

2

σ / 4 ε_{0}

3

σ / ε_{0} R

4

σ / 2 ε_{0}

Explanation:

Electric field at any point on the surface of a thin spherical shell is, $E = \frac{1}{4 π ε_{0}} \cdot \frac{Q}{R^{2}} (1)$
Now, charge on the spherical shell, is,
$Q =$ surface charge density $\times$ Area of spherical shell
supporting img

$∴ Q = σ \times 4 π R^{2}$
Substituting the value of $Q$ in equation (1), we
get $E = \frac{1}{4 π ε_{0}} \times \frac{σ \times 4 π R^{2}}{R^{2}} = \frac{σ}{ε_{0}}$

JEE - 2024

PHXII01:ELECTRIC CHARGES AND FIELDS

358015 A positively charged sphere of radius $r_{0}$ carries a volume charge density $ρ$ . A spherical cavity of radius $r_{0} / 2$ is then scooped out and left empty. $C_{1}$ is the centre of the sphere and $C_{2}$ that of the cavity. What is the direction and magnitude of the electric field at point $B$ ?
supporting img

1

\frac{17 ρ r_{0}}{54 ε_{0}} l e f t

2

\frac{ρ r_{0}}{6 ε_{0}} l e f t

3

\frac{17 ρ r_{0}}{54 ε_{0}} r i g h t

4

\frac{ρ r_{0}}{6 ε_{0}} r i g h t

Explanation:

The electric field at point $B$ is given by
${\vec{E}}_{o r i g i n a l} = {\vec{E}}_{r e m o v e d} + {\vec{E}}_{r e m a i n i n g}$
${\vec{E}}_{o r i g i n a l}$ – Electric field by the original sphere without cavity
${\vec{E}}_{r e m o v e d}$ – Electric field by the removed part
${\vec{E}}_{r e m a i n i n g}$ – Electric field by the remaining part.
${\vec{E}}_{o r i g i n a l} = \frac{- ρ (\frac{4}{3} π r_{0}^{3})}{4 π ε_{0} r_{0}^{2}} \hat{i} = \frac{- ρ r_{0}}{3 ε_{0}} \hat{i}$
${\vec{E}}_{r e m o v e d} = \frac{- ρ \frac{4}{3} π {(\frac{r_{0}}{2})}^{3}}{4 π ε_{0} {(\frac{3 r_{0}}{2})}^{2}} = \frac{- ρ r_{0}}{54 ε_{0}} \hat{i}$
${\vec{E}}_{r e m a i n i n g} = \frac{- 17 ρ r_{0}}{54 ε_{0}} \hat{i}$ (Left to the point $B$ )

PHXII01:ELECTRIC CHARGES AND FIELDS

358016 A conducting sphere of radius $20 c m$ has a known charge. If the electric field at a distance $40 c m$ from the centre of the sphere is $1.2 \times 10^{3} N C^{- 1}$ and points radially inwards. The net charge on the sphere is:

1

- 4.5 \times 10^{- 9} C

2

4.5 \times 10^{9} C

3

- 21.3 \times 10^{- 9} C

4

21.3 \times 10^{9} C

Explanation:

supporting img
$r = 40 cm, \vec{E} = 1.2 \times 10^{3} N / C, \vec{E} = \frac{- K q}{r^{2}}$
$q = \frac{- 1.2 \times 10^{3} \times \frac{40}{100} \times \frac{40}{100}}{9 \times 10^{9}} = - 21.3 \times 10^{- 19} C$

PHXII01:ELECTRIC CHARGES AND FIELDS

358017 A spherical portion has been removed from a solid sphere having a charge distributed uniformly in its volume in the figure. The electric field inside the emptied space is
supporting img

1 Zero everywhere

2 Uniform

3 Non-uniform

4 Zero only at its center

Explanation:

Let $O_{1} a n d O_{2}$ are centres of the sphere and the cavity. Let $P$ is an arbitary point in the spherical cavity.
The electric field inside a solid sphere of total charge $Q$ and radius $R$ at a radial $r$ from the centre is
$E = \frac{Q r}{4 π ε_{0} R^{3}} = \frac{ρ r}{3 ε_{0}} (ρ = \frac{Q}{\frac{4}{3} π R^{3}})$
In vector form $\overset{―}{E} = \frac{ρ \overset{―}{r}}{3 ε_{0}}$
The electric field at $P$ is given as
${\overset{―}{E}}_{o r i g i n a l} = {\overset{―}{E}}_{r e m o v e d} + {\overset{―}{E}}_{r e m a i n i n g} = {\overset{―}{E}}_{r e m o v e d} + {\overset{―}{E}}_{p}$
Where ${\overset{―}{E}}_{P}$ - Electric field at point $P$ because of the remaining part.
${\overset{―}{E}}_{o r i g i n a l} = \frac{ρ \overset{―}{O_{1} P}}{3 ε_{0}}, {\overset{―}{E}}_{r e m o v e d} = \frac{ρ \overset{―}{O_{2} P}}{3 ε_{0}}$
$= \frac{ρ}{3 ε_{0}} (\overset{―}{O_{1} P} - \overset{―}{O_{2} P}) = \frac{ρ \overset{―}{O_{2} O_{2}}}{3 ε_{0}}$
The electric field inside the cavity is uniform and its direction is present along $O_{1} O_{2}$
supporting img

PHXII01:ELECTRIC CHARGES AND FIELDS

358018 $4 \times 10^{10}$ electrons are removed from a neutral metal sphere of diameter $20 c m$ placed in air. The magnitude of the electric field (in $N C^{- 1}$ ) at a distance of 20 $c m$ from its centre is

1

z e r o

2

5760

3

640

4

1440

Explanation:

Number of electrons removed $= 4 \times 10^{10}$
Charge on this sphere $= 4 \times 10^{10} \times 1.6 \times 10^{- 19} C$
$= 6.4 \times 10^{- 19} C = 6.4 n C$
Diameter of Sphere, $D = 20 c m = 0.2 m$
Distance from the centre, $d = 20 c m = 0.2 m$
Electric field at point p, $E = \frac{q}{4 π ε_{0} d^{2}}$
$(\frac{9 \times 10^{9} \times 6.4 \times 10^{- 9}}{{(0.2)}^{2}}) N C^{- 1} = 1440 N C^{- 1}$
supporting img

PHXII01:ELECTRIC CHARGES AND FIELDS

358019 $σ$ is the uniform surface charge density of a thin spherical shell of radius $R$ . The electric field at any point on the surface of the spherical shell is

1

σ / ε_{0}

2

σ / 4 ε_{0}

3

σ / ε_{0} R

4

σ / 2 ε_{0}

Explanation:

Electric field at any point on the surface of a thin spherical shell is, $E = \frac{1}{4 π ε_{0}} \cdot \frac{Q}{R^{2}} (1)$
Now, charge on the spherical shell, is,
$Q =$ surface charge density $\times$ Area of spherical shell
supporting img

$∴ Q = σ \times 4 π R^{2}$
Substituting the value of $Q$ in equation (1), we
get $E = \frac{1}{4 π ε_{0}} \times \frac{σ \times 4 π R^{2}}{R^{2}} = \frac{σ}{ε_{0}}$

JEE - 2024

PHXII01:ELECTRIC CHARGES AND FIELDS

358015 A positively charged sphere of radius $r_{0}$ carries a volume charge density $ρ$ . A spherical cavity of radius $r_{0} / 2$ is then scooped out and left empty. $C_{1}$ is the centre of the sphere and $C_{2}$ that of the cavity. What is the direction and magnitude of the electric field at point $B$ ?
supporting img

1

\frac{17 ρ r_{0}}{54 ε_{0}} l e f t

2

\frac{ρ r_{0}}{6 ε_{0}} l e f t

3

\frac{17 ρ r_{0}}{54 ε_{0}} r i g h t

4

\frac{ρ r_{0}}{6 ε_{0}} r i g h t

Explanation:

The electric field at point $B$ is given by
${\vec{E}}_{o r i g i n a l} = {\vec{E}}_{r e m o v e d} + {\vec{E}}_{r e m a i n i n g}$
${\vec{E}}_{o r i g i n a l}$ – Electric field by the original sphere without cavity
${\vec{E}}_{r e m o v e d}$ – Electric field by the removed part
${\vec{E}}_{r e m a i n i n g}$ – Electric field by the remaining part.
${\vec{E}}_{o r i g i n a l} = \frac{- ρ (\frac{4}{3} π r_{0}^{3})}{4 π ε_{0} r_{0}^{2}} \hat{i} = \frac{- ρ r_{0}}{3 ε_{0}} \hat{i}$
${\vec{E}}_{r e m o v e d} = \frac{- ρ \frac{4}{3} π {(\frac{r_{0}}{2})}^{3}}{4 π ε_{0} {(\frac{3 r_{0}}{2})}^{2}} = \frac{- ρ r_{0}}{54 ε_{0}} \hat{i}$
${\vec{E}}_{r e m a i n i n g} = \frac{- 17 ρ r_{0}}{54 ε_{0}} \hat{i}$ (Left to the point $B$ )

PHXII01:ELECTRIC CHARGES AND FIELDS

358016 A conducting sphere of radius $20 c m$ has a known charge. If the electric field at a distance $40 c m$ from the centre of the sphere is $1.2 \times 10^{3} N C^{- 1}$ and points radially inwards. The net charge on the sphere is:

1

- 4.5 \times 10^{- 9} C

2

4.5 \times 10^{9} C

3

- 21.3 \times 10^{- 9} C

4

21.3 \times 10^{9} C

Explanation:

supporting img
$r = 40 cm, \vec{E} = 1.2 \times 10^{3} N / C, \vec{E} = \frac{- K q}{r^{2}}$
$q = \frac{- 1.2 \times 10^{3} \times \frac{40}{100} \times \frac{40}{100}}{9 \times 10^{9}} = - 21.3 \times 10^{- 19} C$

PHXII01:ELECTRIC CHARGES AND FIELDS

358017 A spherical portion has been removed from a solid sphere having a charge distributed uniformly in its volume in the figure. The electric field inside the emptied space is
supporting img

1 Zero everywhere

2 Uniform

3 Non-uniform

4 Zero only at its center

Explanation:

Let $O_{1} a n d O_{2}$ are centres of the sphere and the cavity. Let $P$ is an arbitary point in the spherical cavity.
The electric field inside a solid sphere of total charge $Q$ and radius $R$ at a radial $r$ from the centre is
$E = \frac{Q r}{4 π ε_{0} R^{3}} = \frac{ρ r}{3 ε_{0}} (ρ = \frac{Q}{\frac{4}{3} π R^{3}})$
In vector form $\overset{―}{E} = \frac{ρ \overset{―}{r}}{3 ε_{0}}$
The electric field at $P$ is given as
${\overset{―}{E}}_{o r i g i n a l} = {\overset{―}{E}}_{r e m o v e d} + {\overset{―}{E}}_{r e m a i n i n g} = {\overset{―}{E}}_{r e m o v e d} + {\overset{―}{E}}_{p}$
Where ${\overset{―}{E}}_{P}$ - Electric field at point $P$ because of the remaining part.
${\overset{―}{E}}_{o r i g i n a l} = \frac{ρ \overset{―}{O_{1} P}}{3 ε_{0}}, {\overset{―}{E}}_{r e m o v e d} = \frac{ρ \overset{―}{O_{2} P}}{3 ε_{0}}$
$= \frac{ρ}{3 ε_{0}} (\overset{―}{O_{1} P} - \overset{―}{O_{2} P}) = \frac{ρ \overset{―}{O_{2} O_{2}}}{3 ε_{0}}$
The electric field inside the cavity is uniform and its direction is present along $O_{1} O_{2}$
supporting img

PHXII01:ELECTRIC CHARGES AND FIELDS

358018 $4 \times 10^{10}$ electrons are removed from a neutral metal sphere of diameter $20 c m$ placed in air. The magnitude of the electric field (in $N C^{- 1}$ ) at a distance of 20 $c m$ from its centre is

1

z e r o

2

5760

3

640

4

1440

Explanation:

Number of electrons removed $= 4 \times 10^{10}$
Charge on this sphere $= 4 \times 10^{10} \times 1.6 \times 10^{- 19} C$
$= 6.4 \times 10^{- 19} C = 6.4 n C$
Diameter of Sphere, $D = 20 c m = 0.2 m$
Distance from the centre, $d = 20 c m = 0.2 m$
Electric field at point p, $E = \frac{q}{4 π ε_{0} d^{2}}$
$(\frac{9 \times 10^{9} \times 6.4 \times 10^{- 9}}{{(0.2)}^{2}}) N C^{- 1} = 1440 N C^{- 1}$
supporting img

PHXII01:ELECTRIC CHARGES AND FIELDS

358019 $σ$ is the uniform surface charge density of a thin spherical shell of radius $R$ . The electric field at any point on the surface of the spherical shell is

1

σ / ε_{0}

2

σ / 4 ε_{0}

3

σ / ε_{0} R

4

σ / 2 ε_{0}

Explanation:

Electric field at any point on the surface of a thin spherical shell is, $E = \frac{1}{4 π ε_{0}} \cdot \frac{Q}{R^{2}} (1)$
Now, charge on the spherical shell, is,
$Q =$ surface charge density $\times$ Area of spherical shell
supporting img

$∴ Q = σ \times 4 π R^{2}$
Substituting the value of $Q$ in equation (1), we
get $E = \frac{1}{4 π ε_{0}} \times \frac{σ \times 4 π R^{2}}{R^{2}} = \frac{σ}{ε_{0}}$

JEE - 2024

PHXII01:ELECTRIC CHARGES AND FIELDS

358015 A positively charged sphere of radius $r_{0}$ carries a volume charge density $ρ$ . A spherical cavity of radius $r_{0} / 2$ is then scooped out and left empty. $C_{1}$ is the centre of the sphere and $C_{2}$ that of the cavity. What is the direction and magnitude of the electric field at point $B$ ?
supporting img

1

\frac{17 ρ r_{0}}{54 ε_{0}} l e f t

2

\frac{ρ r_{0}}{6 ε_{0}} l e f t

3

\frac{17 ρ r_{0}}{54 ε_{0}} r i g h t

4

\frac{ρ r_{0}}{6 ε_{0}} r i g h t

Explanation:

The electric field at point $B$ is given by
${\vec{E}}_{o r i g i n a l} = {\vec{E}}_{r e m o v e d} + {\vec{E}}_{r e m a i n i n g}$
${\vec{E}}_{o r i g i n a l}$ – Electric field by the original sphere without cavity
${\vec{E}}_{r e m o v e d}$ – Electric field by the removed part
${\vec{E}}_{r e m a i n i n g}$ – Electric field by the remaining part.
${\vec{E}}_{o r i g i n a l} = \frac{- ρ (\frac{4}{3} π r_{0}^{3})}{4 π ε_{0} r_{0}^{2}} \hat{i} = \frac{- ρ r_{0}}{3 ε_{0}} \hat{i}$
${\vec{E}}_{r e m o v e d} = \frac{- ρ \frac{4}{3} π {(\frac{r_{0}}{2})}^{3}}{4 π ε_{0} {(\frac{3 r_{0}}{2})}^{2}} = \frac{- ρ r_{0}}{54 ε_{0}} \hat{i}$
${\vec{E}}_{r e m a i n i n g} = \frac{- 17 ρ r_{0}}{54 ε_{0}} \hat{i}$ (Left to the point $B$ )

PHXII01:ELECTRIC CHARGES AND FIELDS

358016 A conducting sphere of radius $20 c m$ has a known charge. If the electric field at a distance $40 c m$ from the centre of the sphere is $1.2 \times 10^{3} N C^{- 1}$ and points radially inwards. The net charge on the sphere is:

1

- 4.5 \times 10^{- 9} C

2

4.5 \times 10^{9} C

3

- 21.3 \times 10^{- 9} C

4

21.3 \times 10^{9} C

Explanation:

supporting img
$r = 40 cm, \vec{E} = 1.2 \times 10^{3} N / C, \vec{E} = \frac{- K q}{r^{2}}$
$q = \frac{- 1.2 \times 10^{3} \times \frac{40}{100} \times \frac{40}{100}}{9 \times 10^{9}} = - 21.3 \times 10^{- 19} C$

PHXII01:ELECTRIC CHARGES AND FIELDS

358017 A spherical portion has been removed from a solid sphere having a charge distributed uniformly in its volume in the figure. The electric field inside the emptied space is
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1 Zero everywhere

2 Uniform

3 Non-uniform

4 Zero only at its center

Explanation:

Let $O_{1} a n d O_{2}$ are centres of the sphere and the cavity. Let $P$ is an arbitary point in the spherical cavity.
The electric field inside a solid sphere of total charge $Q$ and radius $R$ at a radial $r$ from the centre is
$E = \frac{Q r}{4 π ε_{0} R^{3}} = \frac{ρ r}{3 ε_{0}} (ρ = \frac{Q}{\frac{4}{3} π R^{3}})$
In vector form $\overset{―}{E} = \frac{ρ \overset{―}{r}}{3 ε_{0}}$
The electric field at $P$ is given as
${\overset{―}{E}}_{o r i g i n a l} = {\overset{―}{E}}_{r e m o v e d} + {\overset{―}{E}}_{r e m a i n i n g} = {\overset{―}{E}}_{r e m o v e d} + {\overset{―}{E}}_{p}$
Where ${\overset{―}{E}}_{P}$ - Electric field at point $P$ because of the remaining part.
${\overset{―}{E}}_{o r i g i n a l} = \frac{ρ \overset{―}{O_{1} P}}{3 ε_{0}}, {\overset{―}{E}}_{r e m o v e d} = \frac{ρ \overset{―}{O_{2} P}}{3 ε_{0}}$
$= \frac{ρ}{3 ε_{0}} (\overset{―}{O_{1} P} - \overset{―}{O_{2} P}) = \frac{ρ \overset{―}{O_{2} O_{2}}}{3 ε_{0}}$
The electric field inside the cavity is uniform and its direction is present along $O_{1} O_{2}$
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PHXII01:ELECTRIC CHARGES AND FIELDS

358018 $4 \times 10^{10}$ electrons are removed from a neutral metal sphere of diameter $20 c m$ placed in air. The magnitude of the electric field (in $N C^{- 1}$ ) at a distance of 20 $c m$ from its centre is

1

z e r o

2

5760

3

640

4

1440

Explanation:

Number of electrons removed $= 4 \times 10^{10}$
Charge on this sphere $= 4 \times 10^{10} \times 1.6 \times 10^{- 19} C$
$= 6.4 \times 10^{- 19} C = 6.4 n C$
Diameter of Sphere, $D = 20 c m = 0.2 m$
Distance from the centre, $d = 20 c m = 0.2 m$
Electric field at point p, $E = \frac{q}{4 π ε_{0} d^{2}}$
$(\frac{9 \times 10^{9} \times 6.4 \times 10^{- 9}}{{(0.2)}^{2}}) N C^{- 1} = 1440 N C^{- 1}$
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PHXII01:ELECTRIC CHARGES AND FIELDS

358019 $σ$ is the uniform surface charge density of a thin spherical shell of radius $R$ . The electric field at any point on the surface of the spherical shell is

1

σ / ε_{0}

2

σ / 4 ε_{0}

3

σ / ε_{0} R

4

σ / 2 ε_{0}

Explanation:

Electric field at any point on the surface of a thin spherical shell is, $E = \frac{1}{4 π ε_{0}} \cdot \frac{Q}{R^{2}} (1)$
Now, charge on the spherical shell, is,
$Q =$ surface charge density $\times$ Area of spherical shell
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$∴ Q = σ \times 4 π R^{2}$
Substituting the value of $Q$ in equation (1), we
get $E = \frac{1}{4 π ε_{0}} \times \frac{σ \times 4 π R^{2}}{R^{2}} = \frac{σ}{ε_{0}}$

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