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General Strategy:

When we deal with a continuous charges, it is helpful to start with pieces of the body, and use point charge formula. Whenever possible, it usually simplifies calculation if you make use of the symmetry. In the present question, since the field point is in the plane that divides the rod in half, there is a symmetry between the upper half and lower half.

Derivation:

To exploit the symmetry in this situation, we notice two things in this problem: (1) every piece of the ring is same distance from the field point P, and (2) the horizontal component of the electric field from two oppositely placed charges on the ring, as shown in Figure 29.6.5, will cancel out, which means that we need to work out only the vertical component.

To get an idea of what to proceed, let us look at the \(z\) component of the electric field from element of arc length \(ds\text{,}\) say from \(dq_1 = \lambda ds \text{.}\) The magnitude of this electric field is

where \(\sqrt{R^2 + D^2} \) is the direct distance from the \(dq_1\) to the field point P. Now, we need to get its \(z\) component by multiplying with \(\cos\,\theta\text{,}\) where

Therefore, the \(z\) component will be

Now, we notice that as we go around the ring, everything is same for every element. Therefore, the net field will just be \(ds\) replaced by the circumference of the ring. Let us we drop 1 from the subscript since this is the net.

We can write this formula more compactly by replacing \(\lambda\, 2\pi R \) by the total charge \(q\) on the ring, and combining the denominator.

Since this is the only non-zero component, this gives the magnitude of the net field at P.

and direction towards \(+z\) axis if \(\lambda\) is positive and \(-z\) axis if \(\lambda\) is negative. Of course, you can write this in a vector notation as well by using unit vector \(\hat u_z\) that points in the positive \(z\) direction.

(a) The net electric field will be superposition of the two fields, one by each ring. To be safe with signs, we work with the vector notation. In vector notation, the field by one ring will have the form

There will be one term from each ring. The net will be

Here \(q_1 = 2\pi R_1 \lambda_1\) and \(q_2 = 2\pi R_2 \lambda_2\)

(b) when \(q_1 = -q_2\text{,}\) we will have

(a) I will use the formula derived for one ring. When origin is at the center of the ring, the axis is \(z\) axis, and point P is \(z=a\text{,}\) then the electric field would be

\begin{equation*} \vec E = k \dfrac{ q\, a}{ \left( R^2 + a^2 \right)^{3/2} }\ \hat u_z, \end{equation*}

where \(q=2\pi R \lambda\text{,}\) the total charge on the ring. The ring at the bottom is like this. So, let us rename this as \(\vec E_1\text{.}\)

Now, we have the second ring whose center is not at the origin, but it is at \(z=D\text{.}\) Therefore, distance to the field point P from this ring will not be \(a\) but \(D-a\) since P is between the two rings.

Furthermore, since this ring is negatively charged, field at this P by this ring will be pointed up in the positive \(z\) direction.

where \(q\) is same as above and \(D \gt a\text{.}\) The net field at P will be a vector sum of these two fields.

(b) We just set \(a=D/2\) in the formula we obtained in (a).

Let us place the arc symmetrically about \(x\) axis in the \(xy\) plane as shown in Figure 29.6.11. This will have the effect of having \(y\) component of electric field zero by symmetry and we will need to work out only the \(x\) component.

As we did for line and ring, we look at electric field of a small segment and treat it as a point charge. Taking into account the direction of the field as shown in the figure, \(x\) component of the electric field from an element of size \(Rd\theta\) at angle \(\theta\) will be

Since the arc spans from an angle \(-\theta_0\) to \(\theta_0\) with \(\theta_0 = L/2R\) using \(s=R\theta\) formula, we integrate this to get the net electric field at origin.

Since this is the only non-zero component, magnitude of electric field is just the magnitude of this quantity.

and direction away from the arc if \(\lambda\) positive and towards arc if negative.

(a) We imagine dividing up the disk into concentric rings as shown in Figure 29.6.13. Consider one representative ring of radius \(r\) of thickness \(dr\text{.}\) The infintesimal charge \(dq\) on the ring will be \(dq = \sigma\, (2 \pi r dr).\)

The electric field of this ring will have only the \(z\) component nonzero. We can utilize the result of electric field of a ring of charges worked out in Checkpoint 29.6.4. We will get the infinitesimal electric field \(dE_z\) by the ring here as

Now, to obtain the contribution of all such rings on the disk, we will integrate (i.e., sum over) from \(r=0\) to \(r=R\text{,}\) giving us the \(z\) component of the net electricv field at P.

We can write this expression in terms of the total charge on the disk

This gives

(b) To take the limit, let us introduce the variable

This would mean we are interested in the limit \(\epsilon \rightarrow 0\text{.}\) Let us express the answer in (a) in \(\epsilon\) and \(R\) in place of \(D\) and \(R\text{.}\)

Recall from Calculus the Mclaurin series of \(\dfrac{1}{\sqrt{ 1 + \epsilon }} \) as

Keeping only the leading two terms from this series we get

which is the electric field at a distance \(D\) from a point charge \(q\text{.}\)

(c) Let us introduce another symbol for the small parameter.

We will take \(\delta\rightarrow 0\) limit. Writing in \(\delta\) and \(R\)

The expansion now gives

Therefore, we get

which we can write back in \(\sigma\text{,}\) the charge density as

In terms of \(\epsilon_0\text{,}\) the permittivity of vacuum, with \(k = 1/4\pi\epsilon_0\text{,}\) we get

This says that when you get very close to a charged surface, the electric field becomes a constant, independent of the distance to the surface. This turns out to be an important result with many applications.

From the electric field between plates of a parallel plate capacitor we have

where \(\sigma = Q/A\text{.}\) Therefore, we have