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Chapter 4: Q53E (page 716)

Point charges q1 = -4.5 nC and q2 = +4.5 nC are separated by 3.1 mm, forming an electric dipole. (a) Find the electric dipole moment (magnitude and direction). (b) The charges are in a uniform electric field whose direction makes an angle of 36.9° with the line connecting the charges. What is the magnitude of this field if the torque exerted on the dipole has magnitude 7.2 x 10-9 N.m?

Short Answer

Expert verified

(a) The magnitude of the electric dipole moment is 14.0×10−12C⋅mand the direction is from the negative charge to the positive charge.

(b) The magnitude of this field if the torque exerted on the dipole has magnitude is 7.2x10−9N⋅mis856N/C

Step by step solution

01

Direction of the charge

Dipole moment is given by

p = qd

Where q is the charge of the point and d is the distance between the two charges.

p=qd=4.5×10−9C(0.0031m)=14.0×10−12C.m

The direction if the electric dipole is from the negative charge to the positive charge.

02

Electric Field

The torque is given by

τ=Epsin⁡θ

Where is the angle between the electric field and electric dipole and E=electric field

E=τpsin⁡θ=7.2×10−9N⋅m14.0×10−12C⋅msin⁡36.9∘=856N/C

The magnitude of this field if the torque exerted on the dipole has magnitude 7.2 x 10-9 N.m is 856N/C

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