Point charges q 1= 4.80 nC and q 2=+ 4.80 nC are separated by a distance of 3.00

Question

Point charges q1= 4.80 nC andq2=+ 4.80 nC are separated by a distance of 3.00 mm , forming an electric dipole.

Part A

Find the magnitude of the electric dipole moment.

1.44×1011

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Correct

Part B

Find the direction of the electric dipole moment?

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Correct

Part C

The charges are in a uniform electric field whose direction makes an angle of 37.0 with the line connecting the charges. What is the magnitude of this field if the torque exerted on the dipole has magnitude 7.30×109 Nm ?

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Point charges q1= 4.80 nC andq2=+ 4.80 nC are separated by a distance of 3.00 mm , forming an electric dipole.

Part A

Find the magnitude of the electric dipole moment.

p =

1.44×1011

  Cm  

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Correct

Part B

Find the direction of the electric dipole moment?

from q1 to q2 from q2 to q1

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Correct

Part C

The charges are in a uniform electric field whose direction makes an angle of 37.0 with the line connecting the charges. What is the magnitude of this field if the torque exerted on the dipole has magnitude 7.30×109 Nm ?

E = N/C

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Explanation / Answer

Point charges q1=- 4.80nC and q2=+ 4.80nC are separated by a distance of 3mm , forming an electric dipole.

1) The magnitude of the electric dipole moment= p =(4.80*10^-9)( 3*10^-3 ) =1.44*10^-11 C.m

2)The direction of the electric dipole moment is from q1 to q2

3) The charges are in a uniform electric field E whose direction makes an angle of 37degrees with the line connecting the charges.

The torque exerted on the dipole = t =7.30×109 N.m

pEsintheta = torque = t

E = t / psintheta

E = 7.30×109/1.44*10^-11 (sin37)

E = 7.30×109/1.44*10^-11 (0.6018)

E =8.4237*10^2 N/C

the magnitude of this field is E =8.4237*10^2 N/C

the magnitude of this field is E = 842.37 N/C

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