1.27 In a certain region of space, electric field is along the z-direction throughout. The magnitude of electric field is, however, not constant but increases uniformly along the positive z-direction, at the rate of <math><msup><mrow><mrow><mn>10</mn></mrow></mrow><mrow><mrow><mn>5</mn></mrow></mrow></msup></math> N<math><msup><mrow><mrow><mi mathvariant="bold-italic">C</mi></mrow></mrow><mrow><mrow><mo>-</mo><mn>1</mn></mrow></mrow></msup></math>per meter. What are the force and torque experienced by a system having a total dipole moment equal to<math><msup><mrow><mrow><mn>10</mn></mrow></mrow><mrow><mrow><mo>-</mo><mn>7</mn></mrow></mrow></msup></math> Cm in the negative z-direction?

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Ncert Solutions Physics Class 12th Physics Ncert Solutions Class 12th Physics Class 12th Electric Charges and Fields

1.27 In a certain region of space, electric field is along the z-direction throughout. The magnitude of electric field is, however, not constant but increases uniformly along the positive z-direction, at the rate of $10^{5}$ N $C^{- 1}$ per meter. What are the force and torque experienced by a system having a total dipole moment equal to $10^{- 7}$ Cm in the negative z-direction?

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alok kumar singh | Contributor-Level 10

3 months ago

1.27 Dipole moment of the system, p = q $\times d l$ = $10^{- 7}$ Cm

Rate of increase of electric field per unit length, $\frac{d E}{d l}$ = $10^{5}$ N $C^{- 1}$

Force experienced by the system is given by the relation, F = qE = q $\times \frac{d E}{d l} \times d l$

= (q $\times d l) \times \frac{d E}{d l}$ =p $\times \frac{d E}{d l}$ = $- 10^{- 7} \times$ $10^{5}$ = ${- 10}^{- 2}$ N

The force is N in the negative z-direction i.e. opposite to the direction of electric field. Hence, the angle between electric field and dipole moment is 180 $°$
.

Torque ( $τ$ ) is given by the relation, $τ$ = pE $\sin ? 180 °$ = 0

Therefore, the torque experienced by the system is zero.

1.27 Dipole moment of the system, p = q<math><mo>×</mo><mi>d</mi><mi>l</mi></math> = <math><msup><mrow><mrow><mn>10</mn></mrow></mrow><mrow><mrow><mo>-</mo><mn>7</mn></mrow></mrow></msup></math> CmRate of increase of electric field per unit length,<math><mfrac><mrow><mrow><mi>d</mi><mi>E</mi></mrow></mrow><mrow><mrow><mi>d</mi><mi>l</mi></mrow></mrow></mfrac></math> =<math><msup><mrow><mrow><mn>10</mn></mrow></mrow><mrow><mrow><mn>5</mn></mrow></mrow></msup></math> N<math><msup><mrow><mrow><mi>C</mi></mrow></mrow><mrow><mrow><mo>-</mo><mn>1</mn></mrow></mrow></msup></math>  Force experienced by the system is given by the relation, F = qE = q<math><mo>×</mo><mfrac><mrow><mrow><mi>d</mi><mi>E</mi></mrow></mrow><mrow><mrow><mi>d</mi><mi>l</mi></mrow></mrow></mfrac><mo>×</mo><mi>d</mi><mi>l</mi></math>  = (q  <math><mo>×</mo><mi>d</mi><mi>l</mi><mo>)</mo><mo>×</mo><mi></mi><mfrac><mrow><mrow><mi>d</mi><mi>E</mi></mrow></mrow><mrow><mrow><mi>d</mi><mi>l</mi></mrow></mrow></mfrac></math> =p <math><mo>×</mo><mi></mi><mfrac><mrow><mrow><mi>d</mi><mi>E</mi></mrow></mrow><mrow><mrow><mi>d</mi><mi>l</mi></mrow></mrow></mfrac></math> =<math><mo>-</mo><msup><mrow><mrow><mn>10</mn></mrow></mrow><mrow><mrow><mo>-</mo><mn>7</mn></mrow></mrow></msup><mi></mi><mo>×</mo></math> <math><msup><mrow><mrow><mn>10</mn></mrow></mrow><mrow><mrow><mn>5</mn></mrow></mrow></msup></math>= <math><msup><mrow><mrow><mo>-</mo><mn>10</mn></mrow></mrow><mrow><mrow><mo>-</mo><mn>2</mn></mrow></mrow></msup></math> NThe force is  N in the negative z-direction i.e. opposite to the direction of electric field. Hence, the angle between electric field and dipole moment is 180 <math><mo>°</mo></math> .Torque ( <math><mi>τ</mi></math> ) is given by the relation, <math><mi>τ</mi></math> = pE <math><mrow><mrow><mi mathvariant="normal">sin</mi></mrow><mo>?</mo><mrow><mn>180</mn><mo>°</mo></mrow></mrow></math> = 0Therefore, the torque experienced by the system is zero.

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