A dipole of moment 0.5 e nm is placed in a uniform electric field with a magnitude of 8 times 104 N/C. What is the magnitude of the torque on the dipole when?
a) the dipole is parallel to the electric field.
b) the dipole is perpendicular to the electric field.
c) the dipole makes an angle of 30 degrees with the electric field.
d) Find the potential energy of the dipole in the electric field for each case. theta = 0 degrees (parallel).

Respuesta :

Answer with Explanation:

We are given that

Dipole moment=[tex]0.5 e nm=0.5\times 1.6\times 10^{-19}\times 10^{-9}=0.8\times 10^{-28}[/tex]C-m

Because [tex]1 e=1.6\times 10^{-19} C[/tex]

[tex]1 nm=10^{-9} m[/tex]

[tex]a^x\cdot a^y=a^{x+y}[/tex]

Magnitude of electric field,E=[tex]8\times 10^4[/tex] N/C

a.We have to find the magnitude of torque on the dipole when

dipole is parallel to the electric field

i.e[tex]\theta=0[/tex]

We know that

[tex]\tau=PEsin\theta[/tex]

Substitute the values then we get

[tex]\tau=0.8\times 10^{-28}\times 8\times 10^{4} sin0=0[/tex]

Because sin 0=0

[tex]\tau=0[/tex]

b.[tex]\theta=90^{\circ}[/tex]

[tex]\tau=0.8\times 10^{-28}\times 8\times 10^4sin90[/tex]

By using [tex]sin90^{\circ}=1[/tex]

[tex]\tau=6.4\times 10^{-28+4}=6.4\times 10^{-24}Nm[/tex]

[tex]\tau=6.4\times 10^{-24}[/tex]Nm

c.[tex]\theta=30^{\circ}[/tex]

[tex]\tau=0.8\times 10^{-28}\times 8\times 10^4sin 30[/tex]

[tex]\tau=6.4\times 10^{-24}\times \frac{1}{2}[/tex] Nm

By using [tex] sin30^{\circ}=\frac{1}{2}[/tex]

[tex]tau=3.2\times 10^{-24} Nm[/tex]

d.Potential energy, U=[tex]-PEcos\theta[/tex]

[tex]\theta=0[/tex]

[tex]U=-0.8\times 10^{-28}\times 8\times 10^4cos 0=-6.4\times 10^{-28}J[/tex]

Because cos 0degree=1

By having the p, E, and anglёs, we can calculate the τ value and the U values. a) 0 Nm, b) 6.4x10⁻²⁴ Nm, c) 3.2x10⁻²⁴ Nm. d) U = 6.4x10⁻²⁴, U = 0, U = 5.57x10⁻²⁴

First, we need to change the units of the given dip0lё momёnt, p.

The result is

p = 0.8 x 10 ⁻²⁸ Cm

Now we can calculate the t0rquё values.

T0rque, t = p x E x sёn θ

The results are

a) τ = 0 Nm

b) τ = 6.4x10⁻²⁴ Nm

c) τ = 3.2x10⁻²⁴ Nm

d)

p0tёntial energy, U = p x E x c0s θ

  • U = 6.4x10⁻²⁴
  • U = 0
  • U = 5.57x10⁻²⁴

You will find the complete explanation in the attached files

You can learn more about t0rquё and p0tential energy at

https://brainly.com/question/258149

https://brainly.com/question/21288807

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