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Chapter 4: Q4.30P (page 205)

An electric dipole p→, pointing in the ydirection, is placed midwaybetween two large conducting plates, as shown in Fig. 4.33. Each plate makes a small angle θwith respect to the xaxis, and they are maintained at potentials ±V.What is the directionof the net force onp→?(There's nothing to calculate,here, butdo explain your answer qualitatively.)

Short Answer

Expert verified

The net force onp→placed midway between two large conducting plates is towards the right.

Step by step solution

01

Given data

There is a dipole with dipole moment p→pointing in the ydirection and placed midway

between two large conducting plates.

The plate makes asmall angle θwith respect to the xaxis.

The plates maintained at potentials ±V.

02

Determine the direction of force on a dipole

The lines of force and the corresponding electric field are always perpendicular to a conducting surface.Thus, for the setup mentioned in the problem, the lines of follows emerge from the upper plate kept at a positive potential and go into the lower plate kept at a negative potential as follows

The corresponding direction of forces on the positive and negative ends of the dipole are also shown in the above figure.

Thus, the net force is towards the right.

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Most popular questions from this chapter

A point charge qis imbedded at the center of a sphere of linear dielectric material (with susceptibilityχeand radius R).Find the electric field, the polarization, and the bound charge densities,ρb and σb.What is the total bound charge on the surface? Where is the compensating negative bound charge located?

A sphere of radius R carries a polarization

P(r)=kr,

Where k is a constant and r is the vector from the center.

(a) Calculate the bound charges σband ρb.

(b) Find the field inside and outside the sphere.

A point dipole p is imbedded at the center of a sphere of linear dielectric material (with radius R and dielectric constant εr). Find the electric potential inside and outside the sphere.

role="math" localid="1658748385913" [Aanswer:pcosθ4πεr21+2r3R3εr-1εr+2,r≤R:pcosθ4πε0r23εr+2,r≥R]

Suppose you have enough linear dielectric material, of dielectric constant ∈rto half-fill a parallel-plate capacitor (Fig. 4.25). By what fraction is the capacitance increased when you distribute the material as in Fig. 4.25(a)? How about Fig. 4.25(b)? For a given potential difference V between the plates, find E, D, and P , in each region, and the free and bound charge on all surfaces, for both cases.

The space between the plates of a parallel-plate capacitor (Fig. 4.24)

is filled with two slabs of linear dielectric material. Each slab has thickness a, sothe total distance between the plates is 2a. Slab 1 has a dielectric constant of 2, andslab 2 has a dielectric constant of 1.5. The free charge density on the top plate is aand on the bottom plate-σ.

(a) Find the electric displacement Dineach slab.

(b) Find the electric field E in each slab.

(c) Find the polarization P in each slab.

(d) Find the potential difference between the plates.

(e) Find the location and amount of all bound charge.

(f) Now that you know all the charge (free and bound), recalculate the field in eachslab, and confirm your answer to (b).
See all solutions

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