Solved:
Answered by AI, Verified by Human Experts
Point charges q1=− 5.00 nC and q2=+ 5.00 nC are separated by distance 3.30 mm , forming an electric dipole.
Point charges q1=− 5.00 nC and q2=+ 5.00 nC are separated by distance 3.30 mm , forming an electric dipole.A) Find the magnitude of the electric dipole moment. Express your answer in coulomb meters to three significant figures.
B) What is the direction of the electric dipole moment?
from q1 to q2
from q2 to q1
C) The charges are in a uniform electric field whose direction makes an angle 36.1 ∘ with the line connecting the charges. What is the magnitude of this field if the torque exerted on the dipole has magnitude 7.00×10−9 N⋅m ?
Express your answer in newtons per coulomb to three significant figures.
Themagnitudeof theelectric dipole momentis1.65 × 10⁻¹¹ Cm.Thedirectionof theelectric dipole momentis fromq1toq2and this is fromnegativetopositive.Themagnitudeof this field if thetorqueisexerted on thedipoleis7.2003 × 10² N/C.What is an electric dipole?Anelectric dipoleis a pair of opposing chargesq & –qspaced by adistance(d).Theorientationofelectric dipolesinspaceis usually fromnegative charge -qtopositive charge +qby default.From the parameters given:Point charges q₁ = − 5.00 nC and q₂ = + 5.00 nCDistance 2a =3.30 × 10⁻³ m bA)Themagnitudeof theelectric dipole momentis:= q × d= 5 × 10⁻⁹ × 3.3 × 10⁻³= 1.65 × 10⁻¹¹ CmB)Thedirectionof theelectric dipole momentis fromq1toq2and this is fromnegativetopositive.C)Provided that the charges are ina uniform electric fieldwhose direction makes an angle36.1° withthe line connecting thecharges, Then:Themagnitudeof thefieldcan be computed as:E = 7.2003 × 10² N/CLearn more aboutelectric dipolehere:brainly.com/question/14284228...