Point charges q 1= 4.50 nC and q 2=+ 4.50 nC are separated by distance 3.30 mm ,

Question

Point charges q1= 4.50 nC and q2=+ 4.50 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?

C.

The charges are in a uniform electric field whose direction makes an angle 36.8 with the line connecting the charges. What is the magnitude of this field if the torque exerted on the dipole has magnitude 7.50×109 Nm ?

Express your answer in newtons per coulomb to three significant figures.

Explanation / Answer

Electric dipole moment is given by:

p = q*d

q = magnitude of charge = 4.50 nC

d = separation distance between charges = 3.30 mm

So,

p = (4.50*10^-9 C)*(3.30*10^-3 m)

p = 1.485*10^-11 C-m = 1.48*10^-11 C-m

Part B.

Direction of electric dipole moment is from negative to positive charge, which means,

from q1 to q2

Part C.

relation between torque in magnetic field is given by:

torque = pXE

torque = p*E*sin A

Given that

A = angle 36.8 deg with the line connecting the charges

torque = 7.50*10^-9 N-m

So,

E = torque/(p*sin A)

E = 7.50*10^-9/(1.485*10^-11*sin 36.8 deg)

E = 843.12 N/C = 843 N/C

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