1.3  Electric Fields of Finite Objects

1.3.1  Electric Field for a Ring of Charge

Prob. 1
The ring of charge below has a charge +q uniformly spread around a ring of radius R as shown below.

What is the electric field direction and magnitude at the center of the ring?

Solution 1:
In this case no calculation is necessary. If we break the ring into infinitesimal pairs, each with charge dq and arc-length ds, we see that the electric fields from the pairs cancel as long as the members of the pairs are chosen so as to be on opposite sides of the ring as shown in figure 1.9.

Figure 1.9: Electric field at the center of a uniform ring of charge +q.

That is to say, the electric field is zero since each infinitesimal dq we pick around the ring has its field canceled by another dq diametrically opposite it.

1.3.2  Electric Field for a Dipole Ring Charge

Now what is the electric field at point P?

Soln: Now we need to do some work. First note that there are two symmetries at work here. Check the figure below.

We see that for any pair of infinitesimal arc-lengths on the positive half-ring we can choose the elements so that the horizontal (i.e. x) components cancel while the vertical (i.e. y) components add. If we choose pairs of infinitesimal arc-lengths from the negative half-ring in the same way, we see that their field components also add vertically and cancel horizontally. Thus, we can say on the basis of symmetry alone that the net electric field points downward at point P.

With the direction known it is now just a question of calculating the magnitude of the net E field. For each infinitesimal arc-length, we have

dE = k dq R2

(1.3.2.11)

as the magnitude of E field produced at the center. To give each dq a location, we can use the angle w.r.t. the vertical (note that we could use the angle w.r.t. the horizontal just as easily). Then, this angle defines the vertical component of dE at point P.

Therefore, we can find the net field contribution of the pair of infinitesimal arc-lengths as

dEnet = dEy + dEy = 2k dqcosq R2

(1.3.2.12)

Now we take advantage of one of our symmetries to only calculate the vertical components since the horizontal components all add. We can take advantage of the second symmetry when we are done calculating the field due to the top half-ring since the bottom must contribute the same field at point P. Thus, we do the calculation for the top half then multiply by two to get the answer for the entire charge distribution (top half-ring positive and bottom half-ring negative).

To do the calculation for the top half-ring, we need to parameterize dq in terms of theta. That's easy enough in that the linear charge density of the top half-ring is q/(p*R) since p*R is the circumference of a half-ring. The arc-length of dq is R*dq. So, now we are set to do our integration to find the net field due to the top half-ring. Note that, since we are doing pairs of dq's, we only need to integrate from 0 to p/2 (i.e. a quarter-circle). Also note that we will use l for the linear charge density.

Enet = ó õ 2 dEcosq = ó õ 2 k dq R2 cosq = ó õ 2kl(R dq) R2 cosq = ó õ p/2 0  2kq pR2 cosq dq = 2kq pR2 ó õ p/2 0  cosq dq = 2kq pR2 é ê ë sinq ù ú û p/2 0  Enet = 2kq pR2 (1.3.2.13)

1.3.3  Electric Field for a Rectangular Sheet of Charge

Figure 1.13: An infinitesimally thin sheet of charge -dQ  with length l and width w. The electric field is evaluated at point P a distance h  above the midpoint of the sheet.

1.3.3  Electric Field for a Rectangular Sheet of Charge

Figure 1.14: A sheet of charge -Q of length l and width w broken into lines of length l and width dy. Note that we show only two of these lines, both a distance y from the midpoint, to point out the symmetry that proves that the net electric field of the two is directed perpendicular to the plane and toward it from point P.

We define the position of each line as distance y from the center of the sheet and let each have a charge dQ. Then the magnitude of the electric field at point P due to each line is

dE = 2k dQ h Ö l2 + 4h2

(1.3.3.18)

The direction of this field contribution is towards the line. If we consider pairs of lines, each having charge dQ, and located y above or below the center of the sheet, then we can exploit cylindrical symmetry again: each rectangle contributes a field whose component parallel to the sheet cancels the parallel component of the other rectangle. We are left with the perpendicular components of the electric field due to both rectangles adding. Therefore, for two rectangles, a distance y above/below the center of the sheet, the net electric field has magnitude

dEnet = 2·2k dQ h Ö l2 + 4h2 cosf = 4k dQ h Ö l2 + 4h2 · h   ______ Ö y2 + h2 = 4k dQ Ö   l2 + 4h2     ______ Öy2 + h2

(1.3.3.19)

where f represents the angle of the electric field vector with respect to a line perpendicular to the plane of the charge sheet. To integrate over all rectangular charges dQ, we first need to note that

dQ = -Q w dy

(1.3.3.20)

Then the net electric field when we integrate over all pairs of rectangles is directed from point P towards the midpoint of the charged sheet and has magnitude

Enet = ó õ dEnet = ó õ w/2 0  4k Ö l2 + 4h2 · Q w · dy   ______ Öy2 + h2 = 4kQ w Ö l2 + 4h2 ln é ê ë y +   ______ Öy2 + h2   ù ú û w/2 0  Enet = 4kQ w Ö l2 + 4h2 ln é ê ê ê ë w +   _______ Öw2 + 4h2   2h ù ú ú ú û (1.3.3.21)

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