QBManager

Electric Charges and Fields

272155 Two point charges $+ Q$ and $+ q$ are separated by a certain distance. If $+ Q > + q$ then in between the charges the electric field is zero at a point

1 closer to

+ Q

2 closer to

+ q

3 exactly at the mid-point of line segment joining

+ Q

and

+ q

.

4 no where on the line segment joining

+ Q

and

+ q

.

Explanation:

(b) Electric field is directly proportional to the magnitude of charge and inversely proportional to the square of the distance from the charge. Therefore charge $+ Q$ produce a comparatively stronger electric field than $+ q$ which get at cancelled with each other at a point closer to $+ q$ .

NCERT Page-20 / N-16

Electric Charges and Fields

272156 If an electron has an initial velocity in a direction different from that of an electric field, the path of the electron is

1 a straight line

2 a circle

3 an ellipse

4 aparabolà

NCERT Page-21 / N-14

Electric Charges and Fields

272157 In the figure, charge $q$ is placed at origin $ρ$ . When the charge $q$ is displaced from its position the electric field at point $P$ changes

1 at the same time when

q

is displaced.

2 at a time after

\frac{O P}{c}

where

c

is the speed of light.

3 at a time after

\frac{O P c o s θ}{c}

4 at a time after

\frac{O P s i n θ}{c}

Explanation:

(b) The electric field around a charge propagates with the speed of light away from the charge. Therefore the required time $= \frac{d i s t a n c e}{s p e e d} = \frac{O P}{c}$ .

NCERT Page-20 / N-14

Electric Charges and Fields

272158 The electric field intensity just sufficient to balance the earth's gravitational attraction on an electron will be: (given mass, charge of an electron respectively are $9.1 \times 10^{- 31} k g$ and $1.6 \times 10^{- 19} C$ .)

1

- 5.6 \times 10^{- 11} N / C

2

- 1.6 \times 10^{- 19} N / C

3

- 4.8 \times 10^{- 15} N / C

4

- 3.2 \times 10^{- 19} N / C

Explanation:

(a) $- e E = m g, \vec{E} = - \frac{9.1 \times 10^{- 31} \times 10}{1.6 \times 10^{- 19}} = - 5.6 \times 10^{- 11} N / C$

NCERT Page-18/N-14

Electric Charges and Fields

272159 $A B C$ is an equilateral triangle. Charges $+ q$ are placed at each corner as shown in fig. The electric intensity at centre $O$ will be

1

\frac{1}{4 π ϵ_{0}} \frac{q}{r}

.

2

\frac{1}{4 π ϵ_{0}} \frac{q}{r^{2}}

3

\frac{1}{4 π ϵ_{o}} \frac{3 q}{r^{2}}

4 zero

Explanation:

(d) Unit positive charge at $O$ will be repelled equally by three charges at the three corners of triangle. By symmetry, resultant $\vec{E}$ at $O$ would be zero.

NCERT Page-20 / N-16

Electric Charges and Fields

272155 Two point charges $+ Q$ and $+ q$ are separated by a certain distance. If $+ Q > + q$ then in between the charges the electric field is zero at a point

1 closer to

+ Q

2 closer to

+ q

3 exactly at the mid-point of line segment joining

+ Q

and

+ q

.

4 no where on the line segment joining

+ Q

and

+ q

.

Explanation:

(b) Electric field is directly proportional to the magnitude of charge and inversely proportional to the square of the distance from the charge. Therefore charge $+ Q$ produce a comparatively stronger electric field than $+ q$ which get at cancelled with each other at a point closer to $+ q$ .

NCERT Page-20 / N-16

Electric Charges and Fields

272156 If an electron has an initial velocity in a direction different from that of an electric field, the path of the electron is

1 a straight line

2 a circle

3 an ellipse

4 aparabolà

NCERT Page-21 / N-14

Electric Charges and Fields

272157 In the figure, charge $q$ is placed at origin $ρ$ . When the charge $q$ is displaced from its position the electric field at point $P$ changes

1 at the same time when

q

is displaced.

2 at a time after

\frac{O P}{c}

where

c

is the speed of light.

3 at a time after

\frac{O P c o s θ}{c}

4 at a time after

\frac{O P s i n θ}{c}

Explanation:

(b) The electric field around a charge propagates with the speed of light away from the charge. Therefore the required time $= \frac{d i s t a n c e}{s p e e d} = \frac{O P}{c}$ .

NCERT Page-20 / N-14

Electric Charges and Fields

272158 The electric field intensity just sufficient to balance the earth's gravitational attraction on an electron will be: (given mass, charge of an electron respectively are $9.1 \times 10^{- 31} k g$ and $1.6 \times 10^{- 19} C$ .)

1

- 5.6 \times 10^{- 11} N / C

2

- 1.6 \times 10^{- 19} N / C

3

- 4.8 \times 10^{- 15} N / C

4

- 3.2 \times 10^{- 19} N / C

Explanation:

(a) $- e E = m g, \vec{E} = - \frac{9.1 \times 10^{- 31} \times 10}{1.6 \times 10^{- 19}} = - 5.6 \times 10^{- 11} N / C$

NCERT Page-18/N-14

Electric Charges and Fields

272159 $A B C$ is an equilateral triangle. Charges $+ q$ are placed at each corner as shown in fig. The electric intensity at centre $O$ will be

1

\frac{1}{4 π ϵ_{0}} \frac{q}{r}

.

2

\frac{1}{4 π ϵ_{0}} \frac{q}{r^{2}}

3

\frac{1}{4 π ϵ_{o}} \frac{3 q}{r^{2}}

4 zero

Explanation:

(d) Unit positive charge at $O$ will be repelled equally by three charges at the three corners of triangle. By symmetry, resultant $\vec{E}$ at $O$ would be zero.

NCERT Page-20 / N-16

Electric Charges and Fields

272155 Two point charges $+ Q$ and $+ q$ are separated by a certain distance. If $+ Q > + q$ then in between the charges the electric field is zero at a point

1 closer to

+ Q

2 closer to

+ q

3 exactly at the mid-point of line segment joining

+ Q

and

+ q

.

4 no where on the line segment joining

+ Q

and

+ q

.

Explanation:

(b) Electric field is directly proportional to the magnitude of charge and inversely proportional to the square of the distance from the charge. Therefore charge $+ Q$ produce a comparatively stronger electric field than $+ q$ which get at cancelled with each other at a point closer to $+ q$ .

NCERT Page-20 / N-16

Electric Charges and Fields

272156 If an electron has an initial velocity in a direction different from that of an electric field, the path of the electron is

1 a straight line

2 a circle

3 an ellipse

4 aparabolà

NCERT Page-21 / N-14

Electric Charges and Fields

272157 In the figure, charge $q$ is placed at origin $ρ$ . When the charge $q$ is displaced from its position the electric field at point $P$ changes

1 at the same time when

q

is displaced.

2 at a time after

\frac{O P}{c}

where

c

is the speed of light.

3 at a time after

\frac{O P c o s θ}{c}

4 at a time after

\frac{O P s i n θ}{c}

Explanation:

(b) The electric field around a charge propagates with the speed of light away from the charge. Therefore the required time $= \frac{d i s t a n c e}{s p e e d} = \frac{O P}{c}$ .

NCERT Page-20 / N-14

Electric Charges and Fields

272158 The electric field intensity just sufficient to balance the earth's gravitational attraction on an electron will be: (given mass, charge of an electron respectively are $9.1 \times 10^{- 31} k g$ and $1.6 \times 10^{- 19} C$ .)

1

- 5.6 \times 10^{- 11} N / C

2

- 1.6 \times 10^{- 19} N / C

3

- 4.8 \times 10^{- 15} N / C

4

- 3.2 \times 10^{- 19} N / C

Explanation:

(a) $- e E = m g, \vec{E} = - \frac{9.1 \times 10^{- 31} \times 10}{1.6 \times 10^{- 19}} = - 5.6 \times 10^{- 11} N / C$

NCERT Page-18/N-14

Electric Charges and Fields

272159 $A B C$ is an equilateral triangle. Charges $+ q$ are placed at each corner as shown in fig. The electric intensity at centre $O$ will be

1

\frac{1}{4 π ϵ_{0}} \frac{q}{r}

.

2

\frac{1}{4 π ϵ_{0}} \frac{q}{r^{2}}

3

\frac{1}{4 π ϵ_{o}} \frac{3 q}{r^{2}}

4 zero

Explanation:

(d) Unit positive charge at $O$ will be repelled equally by three charges at the three corners of triangle. By symmetry, resultant $\vec{E}$ at $O$ would be zero.

NCERT Page-20 / N-16

Electric Charges and Fields

272155 Two point charges $+ Q$ and $+ q$ are separated by a certain distance. If $+ Q > + q$ then in between the charges the electric field is zero at a point

1 closer to

+ Q

2 closer to

+ q

3 exactly at the mid-point of line segment joining

+ Q

and

+ q

.

4 no where on the line segment joining

+ Q

and

+ q

.

Explanation:

(b) Electric field is directly proportional to the magnitude of charge and inversely proportional to the square of the distance from the charge. Therefore charge $+ Q$ produce a comparatively stronger electric field than $+ q$ which get at cancelled with each other at a point closer to $+ q$ .

NCERT Page-20 / N-16

Electric Charges and Fields

272156 If an electron has an initial velocity in a direction different from that of an electric field, the path of the electron is

1 a straight line

2 a circle

3 an ellipse

4 aparabolà

NCERT Page-21 / N-14

Electric Charges and Fields

272157 In the figure, charge $q$ is placed at origin $ρ$ . When the charge $q$ is displaced from its position the electric field at point $P$ changes

1 at the same time when

q

is displaced.

2 at a time after

\frac{O P}{c}

where

c

is the speed of light.

3 at a time after

\frac{O P c o s θ}{c}

4 at a time after

\frac{O P s i n θ}{c}

Explanation:

(b) The electric field around a charge propagates with the speed of light away from the charge. Therefore the required time $= \frac{d i s t a n c e}{s p e e d} = \frac{O P}{c}$ .

NCERT Page-20 / N-14

Electric Charges and Fields

272158 The electric field intensity just sufficient to balance the earth's gravitational attraction on an electron will be: (given mass, charge of an electron respectively are $9.1 \times 10^{- 31} k g$ and $1.6 \times 10^{- 19} C$ .)

1

- 5.6 \times 10^{- 11} N / C

2

- 1.6 \times 10^{- 19} N / C

3

- 4.8 \times 10^{- 15} N / C

4

- 3.2 \times 10^{- 19} N / C

Explanation:

(a) $- e E = m g, \vec{E} = - \frac{9.1 \times 10^{- 31} \times 10}{1.6 \times 10^{- 19}} = - 5.6 \times 10^{- 11} N / C$

NCERT Page-18/N-14

Electric Charges and Fields

272159 $A B C$ is an equilateral triangle. Charges $+ q$ are placed at each corner as shown in fig. The electric intensity at centre $O$ will be

1

\frac{1}{4 π ϵ_{0}} \frac{q}{r}

.

2

\frac{1}{4 π ϵ_{0}} \frac{q}{r^{2}}

3

\frac{1}{4 π ϵ_{o}} \frac{3 q}{r^{2}}

4 zero

Explanation:

(d) Unit positive charge at $O$ will be repelled equally by three charges at the three corners of triangle. By symmetry, resultant $\vec{E}$ at $O$ would be zero.

NCERT Page-20 / N-16

Electric Charges and Fields

272155 Two point charges $+ Q$ and $+ q$ are separated by a certain distance. If $+ Q > + q$ then in between the charges the electric field is zero at a point

1 closer to

+ Q

2 closer to

+ q

3 exactly at the mid-point of line segment joining

+ Q

and

+ q

.

4 no where on the line segment joining

+ Q

and

+ q

.

Explanation:

(b) Electric field is directly proportional to the magnitude of charge and inversely proportional to the square of the distance from the charge. Therefore charge $+ Q$ produce a comparatively stronger electric field than $+ q$ which get at cancelled with each other at a point closer to $+ q$ .

NCERT Page-20 / N-16

Electric Charges and Fields

272156 If an electron has an initial velocity in a direction different from that of an electric field, the path of the electron is

1 a straight line

2 a circle

3 an ellipse

4 aparabolà

NCERT Page-21 / N-14

Electric Charges and Fields

272157 In the figure, charge $q$ is placed at origin $ρ$ . When the charge $q$ is displaced from its position the electric field at point $P$ changes

1 at the same time when

q

is displaced.

2 at a time after

\frac{O P}{c}

where

c

is the speed of light.

3 at a time after

\frac{O P c o s θ}{c}

4 at a time after

\frac{O P s i n θ}{c}

Explanation:

(b) The electric field around a charge propagates with the speed of light away from the charge. Therefore the required time $= \frac{d i s t a n c e}{s p e e d} = \frac{O P}{c}$ .

NCERT Page-20 / N-14

Electric Charges and Fields

272158 The electric field intensity just sufficient to balance the earth's gravitational attraction on an electron will be: (given mass, charge of an electron respectively are $9.1 \times 10^{- 31} k g$ and $1.6 \times 10^{- 19} C$ .)

1

- 5.6 \times 10^{- 11} N / C

2

- 1.6 \times 10^{- 19} N / C

3

- 4.8 \times 10^{- 15} N / C

4

- 3.2 \times 10^{- 19} N / C

Explanation:

(a) $- e E = m g, \vec{E} = - \frac{9.1 \times 10^{- 31} \times 10}{1.6 \times 10^{- 19}} = - 5.6 \times 10^{- 11} N / C$

NCERT Page-18/N-14

Electric Charges and Fields

272159 $A B C$ is an equilateral triangle. Charges $+ q$ are placed at each corner as shown in fig. The electric intensity at centre $O$ will be

1

\frac{1}{4 π ϵ_{0}} \frac{q}{r}

.

2

\frac{1}{4 π ϵ_{0}} \frac{q}{r^{2}}

3

\frac{1}{4 π ϵ_{o}} \frac{3 q}{r^{2}}

4 zero

Explanation:

(d) Unit positive charge at $O$ will be repelled equally by three charges at the three corners of triangle. By symmetry, resultant $\vec{E}$ at $O$ would be zero.

NCERT Page-20 / N-16