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5.5 Calculating electric fields of charge distributions  (Page 4/12)

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The field of a disk

Find the electric field of a circular thin disk of radius R and uniform charge density at a distance z above the center of the disk ( [link] )

A disk of radius R is shown in the x y plane of an x y z coordinate system. The disk is centered on the origin. A ring, concentric with the disk, of radius r prime and width d r prime is indicated and two small segments on opposite sides of the ring are shaded and labeled as having charge d q. The test point is on the z axis, a distance of z above the center of the disk. The distance from each shaded segment to the test point is r. The electric field contributions, d E, due to the d q charges are shown as arrows in the directions of the associated r vectors. The d E vectors are at an angle of theta to the z axis.
A uniformly charged disk. As in the line charge example, the field above the center of this disk can be calculated by taking advantage of the symmetry of the charge distribution.

Strategy

The electric field for a surface charge is given by

E → ( P ) = 1 4 π ε 0 ∫ surface σ d A r 2 r ^ .

To solve surface charge problems, we break the surface into symmetrical differential “stripes” that match the shape of the surface; here, we’ll use rings, as shown in the figure. Again, by symmetry, the horizontal components cancel and the field is entirely in the vertical ( k ^ ) direction. The vertical component of the electric field is extracted by multiplying by cos θ , so

E → ( P ) = 1 4 π ε 0 ∫ surface σ d A r 2 cos θ k ^ .

As before, we need to rewrite the unknown factors in the integrand in terms of the given quantities. In this case,

d A = 2 π r ′ d r ′ r 2 = r ′ 2 + z 2 cos θ = z ( r ′ 2 + z 2 ) 1 / 2 .

(Please take note of the two different “ r ’s” here; r is the distance from the differential ring of charge to the point P where we wish to determine the field, whereas r ′ is the distance from the center of the disk to the differential ring of charge.) Also, we already performed the polar angle integral in writing down dA .

Solution

Substituting all this in, we get

E → ( P ) = E → ( z ) = 1 4 π ε 0 ∫ 0 R σ ( 2 π r ′ d r ′ ) z ( r ′ 2 + z 2 ) 3 / 2 k ^ = 1 4 π ε 0 ( 2 π σ z ) ( 1 z − 1 R 2 + z 2 ) k ^

or, more simply,

E → ( z ) = 1 4 π ε 0 ( 2 π σ − 2 π σ z R 2 + z 2 ) k ^ .

Significance

Again, it can be shown (via a Taylor expansion) that when z ≫ R , this reduces to

E → ( z ) ≈ 1 4 π ε 0 σ π R 2 z 2 k ^ ,

which is the expression for a point charge Q = σ π R 2 .

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Check Your Understanding How would the above limit change with a uniformly charged rectangle instead of a disk?

The point charge would be Q = σ a b where a and b are the sides of the rectangle but otherwise identical.

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As R → ∞ , [link] reduces to the field of an infinite plane    , which is a flat sheet whose area is much, much greater than its thickness, and also much, much greater than the distance at which the field is to be calculated:

E → = σ 2 ε 0 k ^ .

Note that this field is constant. This surprising result is, again, an artifact of our limit, although one that we will make use of repeatedly in the future. To understand why this happens, imagine being placed above an infinite plane of constant charge. Does the plane look any different if you vary your altitude? No—you still see the plane going off to infinity, no matter how far you are from it. It is important to note that [link] is because we are above the plane. If we were below, the field would point in the − k ^ direction.

The field of two infinite planes

Find the electric field everywhere resulting from two infinite planes with equal but opposite charge densities ( [link] ).

The figure shows two vertically oriented parallel plates A and B separated by a distance d. Plate A is positively charged and B is negatively charged. Electric field lines are parallel between the plates and curved outward at the ends of the plates. A charge q is moved from A to B. The work done W equals q times V sub A B, and the electric field intensity E equals V sub A B over d.
Two charged infinite planes. Note the direction of the electric field.

Strategy

We already know the electric field resulting from a single infinite plane, so we may use the principle of superposition to find the field from two.

Solution

The electric field points away from the positively charged plane and toward the negatively charged plane. Since the σ are equal and opposite, this means that in the region outside of the two planes, the electric fields cancel each other out to zero.

However, in the region between the planes, the electric fields add, and we get

E → = σ ε 0 i ^

for the electric field. The i ^ is because in the figure, the field is pointing in the + x -direction.

Significance

Systems that may be approximated as two infinite planes of this sort provide a useful means of creating uniform electric fields.

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Source:  OpenStax, University physics volume 2. OpenStax CNX. Oct 06, 2016 Download for free at http://cnx.org/content/col12074/1.3
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