Question #286065

1. 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 a magnitude 7.2 × 10−9N ∙

m?


1
Expert's answer
2022-01-10T09:16:07-0500

Answer

Charge

q1=q2=4.5nC.| q_1|=|q_2|=4.5nC.

Distance d=3.1mm

a) electric dipole

P=qd=4.5nC×3.1mm=13.95×1012CmP=qd\\=4.5nC\times3.1mm\\=13.95\times10^{-12}C-m

Direction is charge q1 to q2.


b) now electric field is given

E=τPsinθE=\frac{\tau }{P\sin\theta}

Putting all values

E=7.2×10913.95×1012sin36.9°E=\frac{7.2\times10^{-9}}{13.95\times10^{-12}\sin36.9°}

=720013.95×.596=\frac{7200}{13.95\times.596}

=867.85V/m=867.85V/m



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