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From applying Gauss's law to one sheet we found that electric field by one sheet is \(\sigma/2\epsilon_0\text{.}\) The, by superposition, we get electric field between plates to be

where the unit vector \(\hat u\) is pointed from positive plate to negative plate. Now, we can make use of Eq. (31.7.1) to find the potential difference between plates. Integrating from negative plate to positive plate will mean \(d\vec r\) will be in the opposite direction to \(\hat u\text{,}\) which means the dot product will cancel the negative sign and result int

Since electric field is constant between plates, the potential difference between plates will just be

where \(E\) is the magnitude of the electric field. Therefore,

Let \(x\) represent the distance of point \(P_1\) from the negative plate. Since electric field is constant, integration from negative plate to this point will give

With \(\phi_{-}\) taken as zero reference, we get the potential at this point to be

By applying Gauss's law you should convince yourself that the magnitude of electric field of a charges shell depends on the distance \(r\) from the center of the shell with direction being radial from the center of the sphere.

For directions, we will assume \(Q\) to be positive for our calculations.

For electric potential at \(P_\text{out}\text{,}\) we will perform the integral in Eq. (31.7.2) from \(r=\infty\) to \(r=r_\text{out}\text{.}\) Since vector \(d\vec r = \hat u_r\, dr\) and \(\vec E\) is in \(\hat u_r\) direction, their dot product gives

Therefore, we get the following for the potential

In the present case, we will need to perform integral from \(r=\infty\) to \(r=r_\text{in}\text{.}\) From (a), we already know what to expect fror any point from \(r=\infty\) to \(r=R\text{,}\) the radius. Present integral will then be divided up in two parts: \(r=\infty\) to \(r=R\) and \(r=R\) to \(r=r_\text{in}\text{,}\) the later will be zero since \(E\) in this domain is zero.

This says that electric potential is same throughout the interior of the metal sphere.

Common to (a) and (b): Applying spherical symmetry case of Gauss's law to the charge distribution, you can show that the electric field first. Let \(q\) be tht total charge in the sphere.

Using \(q\) and radial unit vector \(\hat u_r\text{,}\) electric field can be written as follows.

Now, we can integrate from reference at infinity to \(r=r_\text{out}\text{,}\) a point outside the sphere along a radial line. To set up the integral, we note that \(d\vec r = \hat u_r\, dr\) here for the radial path, giving us \(\vec E \cdot d\vec r = E\, dr\text{.}\) Hence, the potential is

That is, from outside, the spherical charge distribution acts like as it the entire charge is located at the center.

(b) Here, we split the integral into two parts since we need to integrate from infinity to a point inside and, according to Eq. (31.7.10), electric field has different expression outside and inside.

By applying Gauss's law for cylindrical symmetry, we can show that magnitude of electric field at a distance \(s\) from the wire is given by the following formula.

Here, the field is in the cylindrical radial direction, as given by the unit vector \(\hat u_s\text{,}\) and the element \(d\vec r\) is also in the \(\hat u_s\) direction. Therefore, we will get their dot product to be

\begin{equation*} \vec E\cdot d\vec r = E\, ds. \end{equation*}

Now, let us integrate this along cylindrical radial line from \(s=s_0\text{,}\) the reference, to the point \(s=s\text{.}\) This will give us the following potential difference with respect to the reference point.

Choosing \(\phi_\text{ref} = 0 \text{,}\) we get

Note that we cannot choose reference distance, \(s_0=\infty\text{,}\) since that would give use zero potential at all points. Typically, we will choose \(s_0=1\) in the unit we express distance \(s\text{.}\) Thus, if we express \(s\) in meters, we will set \(s_0=1\text{ m}\) and write potential at \(s\) as

Numerically, now, with \(s=25\text{ cm}= 0.25\text{ m}\) we get

The potential is positive since our reference point at \(1\text{ m}\) is further away from positive charges than the field point at \(25\text{ cm}\text{.}\)

The situation presented has cylindrical symmetry. Therefore, electric field will have only cylindrical radial dependence. Let \(\lambda\) denote linear charge density on the inner conductor. Then, by Gauss's law we will have

We need to find \(\lambda\) from \(V_0\text{.}\) For that, we look at the potential difference we expect by integrating from negative plate, chosen as reference for potential, to the positive plate we will get

This gives

We use this in the electric field.

Since the potentials are in spherical coordinates, it is best to work in spherical coordinates where \(\vec\nabla\) is

This has derivatives with respect to \(\theta\) and \(\phi\text{.}\) Since we will work with spherical coordinates in this problem, it is better to use symbol \(V\) for potential to avoid confusion.

For (a), we use

for (b), we use

Carrying out the derivatives, we get the answers for (a) and (b) to be