Dipole in a Uniform External Field

For a dipole in a uniform field, the torque magnitude is τ = pE sin θ, while the magnitude of potential energy is |U| = pE|cos θ|. Therefore τ/|U| = |tan θ|. At θ = 45°, sin 45° = cos 45°, so the common factors pE cancel and the ratio is 1. The quantities p and E alone are not ratios, and zero would be incorrect because both sine and cosine are nonzero.

The torque magnitude is τ = pE sin θ, and sin 45° = sin 135° = 1/√2. Thus the torque magnitude remains unchanged. Potential energy is U = −pE cos θ. At 45°, cos θ is positive, so U is negative; at 135°, cos θ is negative, so U becomes positive. Hence only the energy sign reverses, while the torque magnitude stays the same.

By definition, the electric dipole moment points from the negative charge to the positive charge. Reversing this vector changes the angle θ with the field to its supplementary angle, 180° − θ. Since cos(180° − θ) = −cos θ, the calculated energy U = −pE cos θ can acquire the opposite sign. The physical field and the unit of p do not change; the error is in orientation and interpretation.

A freely released dipole rotates under the torque τ = pE sin θ and tends to reduce its potential energy U = −pE cos θ. The stable equilibrium is the parallel orientation, θ = 0°, where U = −pE is minimum. Starting from 120°, the torque acts so that the angle decreases toward alignment with the field. The antiparallel orientation at 180° is an unstable maximum-energy position, so option A is correct.

For a slow rotation, the external work equals the increase in the dipole's potential energy, assuming no change in kinetic energy. The energy is U = −pE cos θ. At 90°, U_i = 0 because cos 90° = 0. At 180°, U_f = −(20)(−1) = +20 J. Hence W_ext = U_f − U_i = 20 − 0 = 20 J. The negative sign would describe work done by the field, not the required external work.

The governing relations are torque τ = pE sin θ and potential energy U = −pE cos θ. A reliable method is therefore to define the dipole moment from the negative charge toward the positive charge, identify the angle between p and E, and then choose sine for torque or cosine with the negative sign for energy. Merely adding numbers, assuming zero energy, or assuming maximum torque ignores the physical conditions.

For a dipole in a uniform electric field, the torque magnitude is τ = pE sin θ. Substituting p = 12 C m, E = 2 N/C, and sin 30° = 1/2 gives τ = 12 × 2 × 1/2 = 12 N m. Thus option A is correct. Option B omits the sine factor, option C uses an incorrect numerical reduction, and option D would apply only for 0° or 180°.

The torque magnitude is τ = pE sin θ. With p = 6 C m, E = 4 N/C, and sin 60° = √3/2, the value is τ = 6 × 4 × √3/2 = 12√3 N m. Therefore option B is correct. Option D is pE without the angular factor, while options A and C result from incorrect handling of √3/2.

For torque, use τ = pE sin θ. Although 150° is obtuse, sin 150° equals sin 30° and is therefore 1/2. Hence τ = 10 × 5 × 1/2 = 25 N m, so option C is correct. The value 50 N m ignores the angle, zero would require an angle of 0° or 180°, and 10 N m does not follow from the formula.

The torque magnitude is τ = pE sin θ. For θ = 120°, sin 120° = sin 60° = √3/2. Therefore τ = (8)(5)(√3/2) = 20√3 N m, making option D correct. Option B is simply pE and ignores the angular factor. Options A and C use incorrect numerical or trigonometric factors, so they do not satisfy the torque equation.

The torque is τ = pE sin θ. Maximum torque occurs when θ = 90°, so sin θ = 1 and τmax = pE. Rearranging gives p = τmax/E = 56/7 = 8 C m. Thus option A is correct. The value 7 confuses the field with the dipole moment, 49 multiplies instead of dividing, and 392 is the product rather than the required quotient.

At maximum torque, the dipole is perpendicular to the field, so sin 90° = 1 and τmax = pE. Solving for the field gives E = τmax/p = 81/9 = 9 N/C. Hence option B is correct. Option A is a division error, 72 comes from subtracting or misusing the data, and 729 is an unjustified product or power.

For a dipole in a uniform electric field, the potential energy is U = −pE cos θ. Here pE = 24 J and θ = 150°, so cos 150° = −√3/2. Therefore U = −24(−√3/2) = +12√3 J. Hence option B is correct. Option A has the wrong sign, option C uses an incorrect trigonometric value, and option D would apply only when the cosine is zero.

The dipole potential energy in a uniform field is U = −pE cos θ. Initially, θ = 0°, so Ui = −40 J. Finally, at 60°, Uf = −40 cos 60° = −20 J. For a slow rotation with no change in kinetic energy, external work equals the increase in potential energy: Wext = Uf − Ui = (−20) − (−40) = 20 J. Thus option B is correct; option A confuses pE with work.

Use U = −pE cos θ for a dipole in a uniform electric field. At the parallel position, Ui = −18 J. At 120°, cos 120° = −1/2, so Uf = −18(−1/2) = +9 J. During slow rotation the external work is the change in potential energy: Wext = Uf − Ui = 9 − (−18) = 27 J. Therefore option C is correct; 9 J is only the final energy.

For a dipole, U = −pE cos θ, and work done by the field is Wfield = −ΔU. At 90°, Ui = −32 cos 90° = 0. At 0°, Uf = −32 cos 0° = −32 J. Hence ΔU = −32 − 0 = −32 J, so Wfield = −(−32) = +32 J. Option D is correct. The field does positive work while the dipole moves toward stable alignment.

The dipole energy is U = −pE cos θ. Unstable equilibrium occurs at 180°, giving Ui = −25 cos 180° = +25 J. Stable equilibrium occurs at 0°, giving Uf = −25 cos 0° = −25 J. Thus ΔU = Uf − Ui = −50 J. Since the field work is Wfield = −ΔU, it equals +50 J. Therefore option A is correct; the energy decrease becomes positive field work.

The potential energy is U = −pE cos θ. At stable equilibrium, θ = 0° and Ui = −31 J. At unstable equilibrium, θ = 180° and Uf = +31 J. For slow movement, the external agent supplies the increase in potential energy, so Wext = Uf − Ui = 31 − (−31) = 62 J. Hence option B is correct. The negative value would represent the field’s work, not the required external work.

The torque magnitude is τ = pE sin θ, while the maximum torque is pE. Thus τ/τmax = sin θ = √2/2, which permits θ = 45° or 135° in the usual 0°–180° range. The potential energy is U = −pE cos θ; for U to be negative, cos θ must be positive. This selects θ = 45°, so option A is correct. At 135°, the energy would be positive.

Because τ = pE sin θ and τmax = pE, the given ratio gives sin θ = √2/2. Within 0°–180°, the possible angles are 45° and 135°. The dipole energy is U = −pE cos θ. Positive energy requires cos θ < 0, which occurs at 135° but not at 45°. Therefore option B is correct. The other listed angles do not even satisfy the torque ratio.

For a dipole in a uniform field, the torque magnitude is |τ| = pE sin θ, while the magnitude of potential energy is |U| = pE|cos θ|. Therefore |τ|/|U| = tan θ when cos θ is positive. At θ = 45°, sin θ = cos θ = √2/2, so the ratio is (pE√2/2)/(pE√2/2) = 1. Hence option B is correct; √2 would result from using an incorrect expression.

Since τ = pE sin θ and τmax = pE, half the maximum torque means sin θ = 1/2. In the range 0°–180°, this gives θ = 30° or 150°. The potential energy is U = −pE cos θ, so positive energy requires cos θ to be negative. That condition is satisfied at 150°, not 30°. Therefore option C is correct; 90° gives maximum torque but zero energy.

For a dipole in a uniform electric field, torque is τ = pE sin θ, so half the maximum torque requires sin θ = 1/2. In the range 0°–180°, this occurs at 30° and 150°. Potential energy is U = −pE cos θ; it is negative when cos θ is positive. That condition selects 30°, so option D is correct. At 150°, the energy would be positive.

The torque relation is τ = pE sin θ, and the maximum torque is pE. Therefore τ/τmax = √3/2 requires sin θ = √3/2, giving 60° or 120° between 0° and 180°. Since U = −pE cos θ must be positive, cos θ must be negative. This is true at 120°, not 60°, so option B is correct. At 90°, torque is maximum rather than √3/2 of maximum.

Using τ = pE sin θ, a torque of √3/2 of the maximum means sin θ = √3/2. The possible angles from 0° to 180° are 60° and 120°. The dipole energy is U = −pE cos θ, so negative energy requires cos θ > 0. Cosine is positive at 60° and negative at 120°, making option A the only correct choice. The other listed angles do not meet both requirements.