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Chapter 22: Q60P (page 657)

A certain electric dipole is placed in a uniform electric field of magnitude40N/C

. Figure 22-63 gives the magnitude t of the torque on the dipole versus the anglebetween field and the dipole moment .The vertical axis scale is set byts=1.00×10−28N.m. What is the magnitude ofp→?

Short Answer

Expert verified

The magnitude of the dipole momentp→ is.2.5×10−30C.m

Step by step solution

01

The given data

  1. Electric field of magnitude,E=40N/C
  2. Vertical axis scale,τs=1.00×10−28Nm
02

Understanding the concept of electric dipole

Using the equation of the torque in the electric dipole, we can get the required magnitude of the torque with the given data.

Formula:

The torque acting on a dipole tends to rotate the dipole: (hence the dipole) into the direction of field,E is given by: τ=−pEsinθ (i)

where,θ is the angle between p and E.

03

Calculation of the dipole moment

From equation (i), consideringθas a variable, we note that it reaches its maximum value when,:

θ=−90o

τmax=pE

Thus, withE=40N/Cand(τmax=1.00×10−28Nmdetermined from the graph), the dipole moment is given using above equation as:

1.00×10−28Nm=p(40N/C)p=2.5×10−30C.m

Hence, the value of the dipole moment is.2.5×10−30C.m

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Most popular questions from this chapter

Two charged particles are fixed to an xaxis: Particle 1of chargeq1=2.1×10−8Cis at positionx=20cmand particle 2 of chargeq2=−4.00q1is at positionx=70cm.At what coordinate on the axis (other than at infinity) is the net electric field produced by the two particles equal to zero?

An electron on the axis of an electric dipole is25nmfrom the center of the dipole.What is the magnitude of the electrostatic force on the electron if the dipole moment is3.6×10−29C⋅m? Assume that25nmis much larger than the separation of the charged particles that form the dipole.

Three particles, each with positive chargeQ, form an equilateral triangle, with each side of length d.What is the magnitude of the electric field produced by the particles at the midpoint of any side?

Question: An alpha particle (the nucleus of a helium atom) has a mass of 6.64×10-27kgand a charge of +2e. What are the (a) magnitude and (b) direction of the electric field that will balance the gravitational force on the particle?

A particle of charge−q1is at the origin of an xaxis. (a) At what location on the axis should a particle of charge−4q1be placed so that the net electric field is zero atx=2.0mmon the axis? (b) If, instead, a particle of charge+4q1is placed at that location, what is the direction (relative to the positive direction of the xaxis) of the net electric field atx=2.0mm?
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