A dipole of moment 0.50 e nm is placed in a uniform electric field with a magnit

Question

A dipole of moment 0.50 e nm is placed in a uniform electric field with a magnitude of 3.0 x 104 N/C. What is the magnitude of the torque on the dipole for each of the following situations? (a) the dipole is aligned with the electric field N·m (b) the dipole is transverse to (perpendicular to) the electric field N·m (c) the dipole makes an angle of 29.5° with the direction of the electric field N·m (d) Defining the potential energy to be zero when the dipole is transverse to the electric field, find the potential energy of the dipole in the electric field for the orientations specified in Parts (a) and (c) -29.5°, U-

Explanation / Answer

Given

dipole with dipolemoment p = 0.50*1.6*10^-19 *10^-9 C m

electric field is E = 3.0*10^4 N/C

we know that when the dipole placed inside the electric field the torque will acts on it which is given by

T = p X E = p E sin theta

T = 0.50*1.6*10^-19 *10^-9 *3.0*10^4 sin theta

a) when the dipole is alligned with the electric field then the angle is zero

so torque T = p*E sin theta

  

T = p*E sin 0 = 0 N.m

b)

when the dipole is transverse to the electric field that is the angle is 90 degrees so

the torque is T = p*E sin theta

  

T = p*E sin 90

T = 0.50*1.6*10^-19 *10^-9 *3.0*10^4 sin 90 N.m

T = 2.4*10^-24 N.m

c) when theta =29.5 degrees with the field then the torque is

T = p*E sin theta

  

T = p*E sin 29.5

T = 0.50*1.6*10^-19 *10^-9 *3.0*10^4 sin 29.5 N.m

T =1.18181654*10^-24 N.m

d) the potential energy is U = - p.E = - p*E cos theta

when dipole is transverse to the field the angle theta =90 degrees ,

U = - p*E cos 90 = 0 J

when theta = 0

U = - p*E cos 0 = -p*E = - 0.50*1.6*10^-19 *10^-9 *3.0*10^4 J = 2.4*10^-24 J

when the angle theta = 29.5 degrees with the field

U = - p*E cos 29.5 = - 0.50*1.6*10^-19 *10^-9 *3.0*10^4 cos 29.5 J = 2.0889*10^-24 J

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