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Chapter 24: Q21P (page 711)

Question: The ammonia molecule NH3 has a permanent electric dipole moment equal to,1.47 D where,1Debye=3.34×10-30C.m. Calculate the electric potential due to an ammonia molecule at a point 52.0nmaway along the axis of the dipole. (Set v = 0at infinity.)

Short Answer

Expert verified

Answer:

The electric potential due to an ammonia molecule is163μV.

Step by step solution

01

The given data

  1. The dipole moment of the ammonia molecule, p = 1.47D where1D=3.34×10-30C.m
  2. The distance of the ammonia molecule, r = 52nm
  3. The electric potential at infinity is V = 0.
02

Understanding the concept of the electric potential

Using the given concept of the electric potential, we can get the value of the potential using the dipole moment of the given ammonia molecule.

Formula:

The electric potential of a body due to the dipole, V=14πε0pr2 (i)

03

Calculation of the electric potential

Now the electric potential at the distance is given using the given data in equation (i) as follows:

V=8.99×1091.47×3.34×10-30(52.0×10-9)2=16.3μV

Hence, the value of the potential is 16.3μV.

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Most popular questions from this chapter

Sphere 1 with radius has positive charge . Sphere 2 with radius is far from sphere 1 and initially uncharged. After the separated spheres are connected with a wire thin enough to retain only negligible charge, (a) is potential of sphere 1 greater than, less than, or equal to potential of sphere 2? What fraction of ends up on (b) sphere 1 and (c) sphere 2? (d) What is the ratio of the surface charge densities of the spheres?

The chocolate crumb mystery. This story begins with Problem 60 in Chapter 23. (a) From the answer to part (a) of that problem, find an expression for the electric potential as a function of the radial distance r from the center of the pipe. (The electric potential is zero on the grounded pipe wall.) (b) For the typical volume charge density ρ=-1.1×10-3C/m3 , what is the difference in the electric potential between the pipe’s center and its inside wall? (The story continues with Problem 60 in Chapter 25.)

Starting from Eq. 24-30, derive an expression for the electric field due to a dipole at a point on the dipole axis.

Two charges q=+2.0μC are fixed a distance d=2.0cmapart (Fig. 24-69).

(a) With V=0at infinity, what is the electric potential at point C?

(b) You bring a third chargeq=+2.0μC from infinity to C. How much work must you do?

(c) What is the potential energy U of the three-charge configuration when the third charge is in place?

In Fig. 24-38, what is the net electric potential at point Pdue to the four particles if V = 0at infinity,q = 5.00 fC, and d = 4.00 cm?
See all solutions

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