Respuesta :
Answer:
(a) The electric dipole moment is equal to 16.5×10^-12 Cm
(b) The electric dipole moment is directed along the dipole axis from the negative charge to the positive charge.
(C) Torque on the dipole is equal to the products of the Electric field vector and the dipole moment
The magnitude is given by
pE×Sin(phi)
Explanation:
The dipole moment is equal to the productof the magnitude of the charge and the distance of separation between the charges.
The detailed solution to the problem is contaimed in the attachment below.
Thank you for reading.


The magnitude of the electric dipole moment is 1.65 × 10⁻¹¹ Cm.
The direction of the electric dipole moment is from q1 to q2 and this is from negative to positive.
The magnitude of this field if the torque is exerted on the dipole is 7.2003 × 10² N/C.
What is an electric dipole?
An electric dipole is a pair of opposing charges q & –q spaced by a distance(d). The orientation of electric dipoles in space is usually from negative charge -q to positive charge +q by default.
From the parameters given:
- Point charges q₁ = − 5.00 nC and q₂ = + 5.00 nC
- Distance 2a = 3.30 × 10⁻³ m b
A)
The magnitude of the electric dipole moment is:
= q × d
= 5 × 10⁻⁹ × 3.3 × 10⁻³
= 1.65 × 10⁻¹¹ Cm
B)
The direction of the electric dipole moment is from q1 to q2 and this is from negative to positive.
C)
Provided that the charges are in a uniform electric field whose direction makes an angle 36.1° with the line connecting the charges, Then:
The magnitude of the field can be computed as:
[tex]\mathbf{E = \dfrac{torque}{\phi sin \theta}}[/tex]
[tex]\mathbf{E = \dfrac{7\times 10^{-9} \ N}{1.65 \times 10^{-11} \ C \times 0.5892 }}[/tex]
E = 7.2003 × 10² N/C
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