QBManager

Electric Charges and Fields

272187 An electric dipole, consisting of two opposite charges of $2 \times 10^{- 6} C$ each separated by a distance $3 c m$ is placed in an electric field of $2 \times 10^{5} N / C$ . Torque acting on the dipole is

1

12 \times 10^{- 1} N - m

2

12 \times 10^{- 2} N - m

3

12 \times 10^{- 3} N - m

4

12 \times 10^{- 4} N - m

Explanation:

(c) Charges (q) $= 2 \times 10^{- 6} C$ , Distance (d) $= 3 c m$ $= 3 \times 10^{- 2} m$ and electric field $(E) = 2 \times 10^{5} N / C$ . Torque $(τ) = q . d$ . $E = (2 \times 10^{- 6}) \times (3 \times 10^{- 2}) \times (2 \times 10^{5}) = 12 \times 10^{- 3} N - m$ .

NCERT Page-28 / N-27

Electric Charges and Fields

272191 If the dipole of moment $2.57 \times 10^{- 17} c m$ is placed into an electric field of magnitude $3.0 \times 10^{4} N / C$ such that the fields lines are aligned at $30^{\circ}$ with the line joining $P$ to the dipole, what torque acts on the dipole?

1

7.7 \times 10^{- 13} N m

2

3.855 \times 10^{- 13} N m

3

3.855 \times 10^{- 15} N m

4

7.7 \times 10^{- 15} N m

Explanation:

(b) $τ = (2.57 \times 10^{- 17} C m) (3.0 \times 10^{4} \frac{N}{C}) (\frac{1}{2})$ $= 3.855 \times 10^{- 13} N m .$

NCERT Page-31 / N-27

Electric Charges and Fields

272192 An electric dipole is placed at an angle of $30^{\circ}$ with an electric field of intensity $2 \times 10^{5} N C^{- 1}$ , It experiences a torque of $4 N m$ . Calculate the charge on the dipole if the dipole length is $2 c m$ .

1

8 m C

2

4 m C

3

8 μ C

4

2 m C

Explanation:

(d) Torque, $\vec{τ} = \vec{p} \times \vec{E} = p E s i n θ$
$4 = p \times 2 \times 10^{5} \times s i n 30^{\circ}$
or, $p = \frac{4}{2 \times 10^{5} \times s i n 30^{\circ}} = 4 \times 10^{- 5} C m$
Dipole moment, $p = q \times l$
$q = \frac{p}{l} = \frac{4 \times 10^{- 5}}{0.02} = 2 \times 10^{- 3} C = 2 m C$

NCERT Page-31 / N-27

Electric Charges and Fields

272193 On decreasing the distance between the two charges of a dipole which is perpendicular to electric field and decreasing the angle between the dipole and electric field, the torque on the dipole

1 increases

2 decreases

3 remains same

4 cannot be predicted

Explanation:

(b) Since $τ = p E s i n θ$ on decreasing the distance between the two charges, and on decreasing angle $θ$ between the dipole and electric field, $s i n θ$ decreases therefore torque decreases.

NCERT Page-31 / N-27

Electric Charges and Fields

272187 An electric dipole, consisting of two opposite charges of $2 \times 10^{- 6} C$ each separated by a distance $3 c m$ is placed in an electric field of $2 \times 10^{5} N / C$ . Torque acting on the dipole is

1

12 \times 10^{- 1} N - m

2

12 \times 10^{- 2} N - m

3

12 \times 10^{- 3} N - m

4

12 \times 10^{- 4} N - m

Explanation:

(c) Charges (q) $= 2 \times 10^{- 6} C$ , Distance (d) $= 3 c m$ $= 3 \times 10^{- 2} m$ and electric field $(E) = 2 \times 10^{5} N / C$ . Torque $(τ) = q . d$ . $E = (2 \times 10^{- 6}) \times (3 \times 10^{- 2}) \times (2 \times 10^{5}) = 12 \times 10^{- 3} N - m$ .

NCERT Page-28 / N-27

Electric Charges and Fields

272191 If the dipole of moment $2.57 \times 10^{- 17} c m$ is placed into an electric field of magnitude $3.0 \times 10^{4} N / C$ such that the fields lines are aligned at $30^{\circ}$ with the line joining $P$ to the dipole, what torque acts on the dipole?

1

7.7 \times 10^{- 13} N m

2

3.855 \times 10^{- 13} N m

3

3.855 \times 10^{- 15} N m

4

7.7 \times 10^{- 15} N m

Explanation:

(b) $τ = (2.57 \times 10^{- 17} C m) (3.0 \times 10^{4} \frac{N}{C}) (\frac{1}{2})$ $= 3.855 \times 10^{- 13} N m .$

NCERT Page-31 / N-27

Electric Charges and Fields

272192 An electric dipole is placed at an angle of $30^{\circ}$ with an electric field of intensity $2 \times 10^{5} N C^{- 1}$ , It experiences a torque of $4 N m$ . Calculate the charge on the dipole if the dipole length is $2 c m$ .

1

8 m C

2

4 m C

3

8 μ C

4

2 m C

Explanation:

(d) Torque, $\vec{τ} = \vec{p} \times \vec{E} = p E s i n θ$
$4 = p \times 2 \times 10^{5} \times s i n 30^{\circ}$
or, $p = \frac{4}{2 \times 10^{5} \times s i n 30^{\circ}} = 4 \times 10^{- 5} C m$
Dipole moment, $p = q \times l$
$q = \frac{p}{l} = \frac{4 \times 10^{- 5}}{0.02} = 2 \times 10^{- 3} C = 2 m C$

NCERT Page-31 / N-27

Electric Charges and Fields

272193 On decreasing the distance between the two charges of a dipole which is perpendicular to electric field and decreasing the angle between the dipole and electric field, the torque on the dipole

1 increases

2 decreases

3 remains same

4 cannot be predicted

Explanation:

(b) Since $τ = p E s i n θ$ on decreasing the distance between the two charges, and on decreasing angle $θ$ between the dipole and electric field, $s i n θ$ decreases therefore torque decreases.

NCERT Page-31 / N-27

Electric Charges and Fields

272187 An electric dipole, consisting of two opposite charges of $2 \times 10^{- 6} C$ each separated by a distance $3 c m$ is placed in an electric field of $2 \times 10^{5} N / C$ . Torque acting on the dipole is

1

12 \times 10^{- 1} N - m

2

12 \times 10^{- 2} N - m

3

12 \times 10^{- 3} N - m

4

12 \times 10^{- 4} N - m

Explanation:

(c) Charges (q) $= 2 \times 10^{- 6} C$ , Distance (d) $= 3 c m$ $= 3 \times 10^{- 2} m$ and electric field $(E) = 2 \times 10^{5} N / C$ . Torque $(τ) = q . d$ . $E = (2 \times 10^{- 6}) \times (3 \times 10^{- 2}) \times (2 \times 10^{5}) = 12 \times 10^{- 3} N - m$ .

NCERT Page-28 / N-27

Electric Charges and Fields

272191 If the dipole of moment $2.57 \times 10^{- 17} c m$ is placed into an electric field of magnitude $3.0 \times 10^{4} N / C$ such that the fields lines are aligned at $30^{\circ}$ with the line joining $P$ to the dipole, what torque acts on the dipole?

1

7.7 \times 10^{- 13} N m

2

3.855 \times 10^{- 13} N m

3

3.855 \times 10^{- 15} N m

4

7.7 \times 10^{- 15} N m

Explanation:

(b) $τ = (2.57 \times 10^{- 17} C m) (3.0 \times 10^{4} \frac{N}{C}) (\frac{1}{2})$ $= 3.855 \times 10^{- 13} N m .$

NCERT Page-31 / N-27

Electric Charges and Fields

272192 An electric dipole is placed at an angle of $30^{\circ}$ with an electric field of intensity $2 \times 10^{5} N C^{- 1}$ , It experiences a torque of $4 N m$ . Calculate the charge on the dipole if the dipole length is $2 c m$ .

1

8 m C

2

4 m C

3

8 μ C

4

2 m C

Explanation:

(d) Torque, $\vec{τ} = \vec{p} \times \vec{E} = p E s i n θ$
$4 = p \times 2 \times 10^{5} \times s i n 30^{\circ}$
or, $p = \frac{4}{2 \times 10^{5} \times s i n 30^{\circ}} = 4 \times 10^{- 5} C m$
Dipole moment, $p = q \times l$
$q = \frac{p}{l} = \frac{4 \times 10^{- 5}}{0.02} = 2 \times 10^{- 3} C = 2 m C$

NCERT Page-31 / N-27

Electric Charges and Fields

272193 On decreasing the distance between the two charges of a dipole which is perpendicular to electric field and decreasing the angle between the dipole and electric field, the torque on the dipole

1 increases

2 decreases

3 remains same

4 cannot be predicted

Explanation:

(b) Since $τ = p E s i n θ$ on decreasing the distance between the two charges, and on decreasing angle $θ$ between the dipole and electric field, $s i n θ$ decreases therefore torque decreases.

NCERT Page-31 / N-27

Electric Charges and Fields

272187 An electric dipole, consisting of two opposite charges of $2 \times 10^{- 6} C$ each separated by a distance $3 c m$ is placed in an electric field of $2 \times 10^{5} N / C$ . Torque acting on the dipole is

1

12 \times 10^{- 1} N - m

2

12 \times 10^{- 2} N - m

3

12 \times 10^{- 3} N - m

4

12 \times 10^{- 4} N - m

Explanation:

(c) Charges (q) $= 2 \times 10^{- 6} C$ , Distance (d) $= 3 c m$ $= 3 \times 10^{- 2} m$ and electric field $(E) = 2 \times 10^{5} N / C$ . Torque $(τ) = q . d$ . $E = (2 \times 10^{- 6}) \times (3 \times 10^{- 2}) \times (2 \times 10^{5}) = 12 \times 10^{- 3} N - m$ .

NCERT Page-28 / N-27

Electric Charges and Fields

272191 If the dipole of moment $2.57 \times 10^{- 17} c m$ is placed into an electric field of magnitude $3.0 \times 10^{4} N / C$ such that the fields lines are aligned at $30^{\circ}$ with the line joining $P$ to the dipole, what torque acts on the dipole?

1

7.7 \times 10^{- 13} N m

2

3.855 \times 10^{- 13} N m

3

3.855 \times 10^{- 15} N m

4

7.7 \times 10^{- 15} N m

Explanation:

(b) $τ = (2.57 \times 10^{- 17} C m) (3.0 \times 10^{4} \frac{N}{C}) (\frac{1}{2})$ $= 3.855 \times 10^{- 13} N m .$

NCERT Page-31 / N-27

Electric Charges and Fields

272192 An electric dipole is placed at an angle of $30^{\circ}$ with an electric field of intensity $2 \times 10^{5} N C^{- 1}$ , It experiences a torque of $4 N m$ . Calculate the charge on the dipole if the dipole length is $2 c m$ .

1

8 m C

2

4 m C

3

8 μ C

4

2 m C

Explanation:

(d) Torque, $\vec{τ} = \vec{p} \times \vec{E} = p E s i n θ$
$4 = p \times 2 \times 10^{5} \times s i n 30^{\circ}$
or, $p = \frac{4}{2 \times 10^{5} \times s i n 30^{\circ}} = 4 \times 10^{- 5} C m$
Dipole moment, $p = q \times l$
$q = \frac{p}{l} = \frac{4 \times 10^{- 5}}{0.02} = 2 \times 10^{- 3} C = 2 m C$

NCERT Page-31 / N-27

Electric Charges and Fields

272193 On decreasing the distance between the two charges of a dipole which is perpendicular to electric field and decreasing the angle between the dipole and electric field, the torque on the dipole

1 increases

2 decreases

3 remains same

4 cannot be predicted

Explanation:

(b) Since $τ = p E s i n θ$ on decreasing the distance between the two charges, and on decreasing angle $θ$ between the dipole and electric field, $s i n θ$ decreases therefore torque decreases.

NCERT Page-31 / N-27