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The electric force due to a uniform external electric field causes a torque of magnitude 20.0 × 10−9 N⋅m on an electric dipole oriented at 30∘ from the direction of the external field. The dipole moment of the dipole is 9.0 × 10−12 C⋅m.

What is the magnitude of the external electric field?

If the two particles that make up the dipole are 2.5 mm apart, what is the magnitude of the charge on each particle?

Respuesta :

[tex]\bull\sf Torque=\tau=20.0\times 10^{-9}Nm[/tex]

[tex]\bull\sf \theta=30°[/tex]

[tex]\bull\sf E=9.0\times C.m=9.0\times 10^{-12}N/C[/tex]

Now we know

[tex]\\ \sf\longmapsto \tau=pEsin\theta[/tex]

[tex]\\ \sf\longmapsto p=\dfrac{\tau}{Esin\theta}[/tex]

[tex]\\ \sf\longmapsto p=\dfrac{20\times 10^{-9}}{9\times 10^{-12}\times sin30}[/tex]

[tex]\\ \sf\longmapsto p=\dfrac{20\times 10^{-9}}{9\times 10^{-12}\times \dfrac{1}{2}}[/tex]

[tex]\\ \sf\longmapsto p=\dfrac{20\times 10^{-9}}{9\times 10^{-12}}\times 2[/tex]

[tex]\\ \sf\longmapsto p=4.4\times 10^3N/C[/tex]

onyebuchinnaji

The magnitude of the externla electric field is 4,444.44 N/C.

The magnitude of the charge is 3.1 x 10⁻¹² C.

Magnitude of the external electric field

The magnitude of the externla electric field is calculated as follows;

EPsinθ = τ

where;

  • E is the external electric field
  • P is dipole moment
  • τ is torque

[tex]E = \frac{\tau}{P sin(\theta)} \\\\E = \frac{20 \times 10^{-9}}{9\times 10^{-12} \times sin(30)} \\\\E = 4,444.44 \ N/C[/tex]

Magnitude of the charge

The magnitude of the charge is calculated as follows;

[tex]E = \frac{kQ}{r^2} \\\\Q = \frac{Er^2}{k} \\\\Q = \frac{(4,444.44) \times (2.5 \times 10^{-3})^2}{(9\times 10^9)} \\\\Q = 3.1 \times 10^{-12} \ C[/tex]

Learn more about electric field here: https://brainly.com/question/14372859

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