Point charges q1= - 4.30 nC and q2=+ 4.30 nC are separated by a distance of 3.00 mm, forming an electric dipole.
*Find the magnitude of the electric dipole moment. (C/m)
*Find the direction of the electric dipole moment? ( From q1 to q2 or From q2 to q1??)
*The charges are in a uniform electric field whose direction makes an angle of 36.9 degrees with the line connecting the charges. What is the magnitude of this field if the torque exerted on the dipole has magnitude 7.20×10^-9 N*m ? (N/C)

Respuesta :

Answer:

a) p = 25.8 10⁻¹² C m , b) The direction of the dipole moment is directed from the negative to the positive charge, c)  E = 4.65  10² N/C  

Explanation:

a) The dipole moment is

        p = 2qa

        p = 2 4.30 10⁻⁹ 3.00 10⁻³

        p = 25.8 10⁻¹² C m

b) The direction of the dipole moment is directed from the negative to the positive charge, that is, in the opposite direction to the electric field.

c) The torque is

          τ = p x E

          τ = p E sin  θ

          E = τ / p sin θ

          E = 7.20 10⁻⁹ /(25.8 10⁻¹²  sin 36.9)

          E = 4.65  10² N/C            

We calculate

           τ = 15.49 10⁻¹² 4.7 10²

           τ = 7.28 10⁻⁹ N m

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