Sphere with Cavities!

a) Visualizing the symmetry of the system will guide you toward the answer.

b) Use Gauss’s Law to determine the electric field, and then use the principle of superposition.

a) Due to the spherical symmetry and the uniformity of the charge distribution, the electric field at the origin is zero.

b) Gauss’s Law states:

\begin{equation*}
\oint_S \vec{E}\cdot d \vec{A}=\frac{Q_{\text{enc}}}{\epsilon_0}.
\end{equation*}

Consider three spheres: a positive sphere with radius \(2a\) and two negative spheres with radius \(a\).

By Gauss’ law, the electric field of each sphere is:

\begin{equation*}
\vec{E}=\frac{Q}{4\pi\epsilon_0 r^2}\,\hat{\textbf{r}},
\end{equation*}

where \(Q= \frac{ 4 }{3}\rho \pi r^3\). By the principle of superposition, the electric field at \( (0,2a) \) will be the sum along the \({y-}\)-axis of the three spheres. Then,

\begin{equation*}
\vec{E}_{\text{total}}=\frac{\rho\frac{4\pi(2a)^3}{3}}{4\pi\epsilon_0 r_1^2}\,\hat{\textbf{j}}+\frac{-\rho\frac{4\pi(a)^3}{3}}{4\pi\epsilon_0 r_2^2}\,\hat{\textbf{j}}+\frac{\rho\frac{4\pi(a)^3}{3}}{4\pi\epsilon_0 r_3^2}\,\hat{\textbf{j}}.
\end{equation*}

After some simplification, we get:

\begin{equation*}
\vec{E}_{\text{total}}=\frac{8\rho a}{27\epsilon_0}\,\hat{\textbf{j}},
\end{equation*}

which, with numerical values, is:

\begin{equation*}
\vec{E}_{\text{total}}\approx 2.68\times 10^{5}\,\text{N/C}\,\hat{\textbf{j}}.
\end{equation*}

For a more detailed explanation of any of these steps, click on “Detailed Solution”.

a) Due to the spherical symmetry and the uniformity of the charge distribution, the electric field at the origin is zero.

b) To calculate the electric field at point \((0,2a)\), we must use the superposition principle and some simplifications for uniform charge distributions. First, we can imagine the cavities of radius \(a\) as a superposition of positive and negative charges. They will cancel out, and, as a result, we’ll have zero charges. Figure 1 illustrates this concept better. We can see that the given sphere can be modeled as the superposition of a solid sphere of radius \(2a\) and charge density \(\rho\) and two solid spheres of radius \(a\) and charge density \(-\rho\).

Figure 1: The given insulating sphere with holes can be modeled using the superposition principle as the sum of a positive sphere with electric charge density \(\rho\) and radius \(2a\) and two negative spheres with electric charge density \(-\rho\) and radius \(a\).

Notice that we want the electric field on the upper border of the charge array. According to the decomposition shown in the previous figure, it is also on the border of the positively charged sphere and on the border of the upper negative sphere. We’ll now show that for points in the border and outside a uniformly charged sphere, the electric field is the same as the one for a point charge located at the center of the distribution.

For all the charge distributions, we’ll use Gauss’ law to find the electric field \(\vec{E}\):

\begin{equation}
\label{gauss}
\oint_S \vec{E}\cdot d \vec{A}=\frac{Q_{\text{enc}}}{\epsilon_0},
\end{equation}

where the left hand side is the integral for the electric flux through the Gaussian surface \(S\), \(d\vec{A}\) is a vector whose magnitude is the differential surface area \(dA\) and its direction is perpendicular to the surface \(S\). On the right-hand side of equation \eqref{gauss}, we find the term \(Q_{\text{enc}}\), which is the enclosed charge by the surface \(S\) and \(\epsilon_0\) is a physical constant known as the permittivity of free space.

Because in the three charge distributions our Gaussian surfaces will be spherical, we can develop the left-hand side of equation \eqref{gauss} to obtain a general expression for any spherically symmetric charge configuration. Due to the spherical symmetry, the value of the electric field at the spherical surface \(S\) has a constant magnitude and if the charge enclosed is positive, the field will be radially outwards. If the enclosed charge is negative, the electric field will be radially inwards. We’ll develop the case of positive charge, but the procedure and arguments apply also for a negative charge. In the case of a positive charge, the electric field \(\vec{E}\) will be directed radially outwards. Then, we can write

\begin{equation}
\label{efield}
\vec{E}=E\,\hat{\textbf{r}},
\end{equation}

where \(E\) is the magnitude of the electric field and \(\hat{\textbf{r}}\) is the unitary vector in the radial direction.

In the case of the sphere, vector \(d\vec{A}\) can be written as

\begin{equation}
\label{da}
d\vec{A}=dA\,\hat{\textbf{r}}.
\end{equation}

Notice that the vector \(\hat{\textbf{r}}\) is a unitary vector and is perpendicular to the spherical surface \(S\). Thus, using equations \eqref{efield} and \eqref{da}, we can write equation \eqref{gauss} for our particular case as

\begin{equation}
\label{gauss2}
\oint_{S} (\vec{E}\,\hat{\textbf{r}})\cdot (dA\,\hat{\textbf{r}}) =\frac{Q_\text{enc}}{\epsilon_0}.
\end{equation}

Now, we can perform the dot product inside the integral of equation \eqref{gauss2}; remember that the dot product of two vectors \(\vec{a}\) and \(\vec{b}\) can be calculated as \(\vec{a}\cdot\vec{b}=ab\,\cos(\phi)\), where \(a\) and \(b\) are the magnitudes of the vectors and \(\phi\) is the angle between them. In our case, both vectors go in the same direction, so the angle between them is zero; then,

\begin{equation}
\label{dotprod}
(E\,\hat{\textbf{r}})\cdot(dA\,\hat{\textbf{r}})=E dA \cos(0)= E dA.
\end{equation}

Using equation \eqref{dotprod} into equation \eqref{gauss2}, we can simply write

\begin{equation}
\label{gauss4}
\oint_S E_\circ dA=\frac{Q_\text{enc}}{\epsilon_0}.
\end{equation}

Notice that the magnitude of the electric field of the sphere \(E\) is constant along the whole surface \(S\), so we can take it out of the integral
\begin{equation}
\label{gauss5}
E\oint_S dA=\frac{Q_\text{enc}}{\epsilon_0},
\end{equation}

and perform the integral, which gives us the surface area of \(S\), a quantity that we denote by \(A_S\)

\begin{equation}
\label{area}
\oint_S dA=A_S.
\end{equation}

Remember that \(S\) is a sphere of radius \(r\), so \(A_S=4\pi r^2\) and equation \eqref{gauss5} (together with \eqref{area}) becomes

\begin{equation}
\label{gaussesfera}
E(4\pi r^2)=\frac{Q_\text{enc}}{\epsilon_0},
\end{equation}

For each charge distribution, we make a spherical Gaussian surface just at the border of each charge distribution. The whole charge \(Q\) of the insulator is enclosed by the Gaussian surface \(S\), as seen in figure 2.

Figure 2: Spherical Gaussian surfaces (grey dashed line) for the array of charges. For the sphere on the left, the Gaussian surface has radius \(2a\). For the spheres on the right, the Gaussian surfaces have a radius \(a\).

This implies that

\begin{equation}
\label{q3}
Q_{\text{enc}}=Q.
\end{equation}
In order to better appreciate that the electric field is either parallel or anti-parallel to the area differential for these Gaussian spheres, consider figure 3 and figure 4 (one for the positively charged sphere, and the other one for the negatively charged):

Figure 3: Spherical Gaussian surface (pink) enclosing the positively charged sphere. Because the charge is positive the electric field lines \(\vec{E}\) come out of the sphere. The surface area differential \(d\vec{A}\) is perpendicular to the Gaussian surface and thus parallel to the electric field.

Figure 4: Spherical Gaussian surface (pink) enclosing the negatively charged sphere. Because the charge is positive, the electric field lines \(\vec{E}\) point towards the center of the sphere. The surface area differential \(d\vec{A}\) is perpendicular to the Gaussian surface and thus antiparallel (at an \(180^{\circ}\) angle) to the electric field.

Using equation \eqref{gaussesfera} and the result for \(Q_{\text{enc}}\) of equation \eqref{q3}, we obtain

\begin{equation}
E(4\pi r^2)=\frac{Q}{\epsilon_0},
\end{equation}

which can be solved for \(E\) to get

\begin{equation}
E=\frac{Q}{4\pi\epsilon_0 r^2},
\end{equation}

or in vector form (according to equation \eqref{efield})

\begin{equation}
\label{e3}
\vec{E}=\frac{Q}{4\pi\epsilon_0 r^2}\,\hat{\textbf{r}},
\end{equation}

which is equal to the expression for the electric field of a point charge \(Q\) located at the center of the sphere.

To further simplify our problem, we can find the equivalent charge \(Q\) for the positive sphere and for the two negative spheres.

For the positive charge, we’ll use the relation

\begin{equation}
\label{qmas}
Q=\rho V,
\end{equation}

where \(V\) is the volume of the positively charged sphere, which is given by the expression

\begin{equation}
\label{volmas}
V=\frac{4\pi(2a)^3}{3}.
\end{equation}

Using the result of equation \eqref{volmas} into equation \eqref{qmas}, we get that the equivalent point charge is

\begin{equation}
\label{qmas2}
Q=\rho \frac{4\pi (2a)^3}{3}.
\end{equation}

For the negative charges, we have that \(Q\) is given by

\begin{equation}
\label{qmenos}
Q=-\rho V,
\end{equation}

with volume equal to

\begin{equation}
V=\frac{4\pi a^3}{3}.
\end{equation}

Using the expression above in equation \eqref{qmenos}, we get

\begin{equation}
\label{qmenos2}
Q=-\rho \frac{4\pi a^{3}}{3}.
\end{equation}

Then, we can reduce the array shown in figure 2 to the one shown in figure 5.

Figure 5: The initial array is reduced to the superposition of one positive point charge and two negative point charges located at the center of the spheres. This approximation is only valid for electric field calculations at distances equal or greater than \(2a\) from the origin.

Now, we can find the electric field \(\vec{E}_{\text{total}}\) at point \((0,2a)\) by adding up the electric field produced by the three point charges seen in the previous figure. Namely,

\begin{equation}
\vec{E}_{\textbf{total}}=\vec{E}_1+\vec{E}_2+\vec{E}_3,
\end{equation}

where \(\vec{E}_1\) corresponds to the electric field due to the positive charge, \(\vec{E}_2\) corresponds to one due to the negative charge above the X axis and \(\vec{E}_3\) corresponds to the electric field of the negative charge below the X axis. Using the definition in equation \eqref{e3} and the expression for the positive and negative charges given by equations \eqref{qmas2} and \eqref{qmenos2}, we have:

\begin{equation}
\label{etot}
\vec{E}_{\text{total}}=\frac{\rho\frac{4\pi(2a)^3}{3}}{4\pi\epsilon_0 r_1^2}\,\hat{\textbf{r}}_1+\frac{-\rho\frac{4\pi(a)^3}{3}}{4\pi\epsilon_0 r_2^2}\,\hat{\textbf{r}}_2+\frac{\rho\frac{4\pi(a)^3}{3}}{4\pi\epsilon_0 r_3^2}\,\hat{\textbf{r}}_3,
\end{equation}

where \(r_1\), \(r_2\) and \(r_3\) are the distances from the point charges to the point \((0,2a)\). In all cases, the unitary vector \(\hat{\textbf{r}}_1=\hat{\textbf{r}}_2=\hat{\textbf{r}}_3=\hat{\textbf{j}}\) because all the charges and the point of interest are located over the Y axis. We can then simplify the expression in \eqref{etot} to

\begin{equation}
\label{etotal2}
\vec{E}_{\text{total}}=\frac{\rho (2a)^3}{3\epsilon_0 r_1^2}\,\hat{\textbf{j}}-\frac{\rho (a)^3}{3\epsilon_0 r_2^2}\,\hat{\textbf{j}}-\frac{\rho(a)^3}{3\epsilon_0r_3^2}\,\hat{\textbf{j}}.
\end{equation}

From figure 5, it is clear that \(r_1=2a\), \(r_2=a\) and \(r_3=3a\). Then, equation \eqref{etotal2} becomes

\begin{equation}
\label{etotal3}
\vec{E}_{\text{total}}=\frac{\rho (2a)^3}{3\epsilon_0 (2a)^2}\,\hat{\textbf{j}}-\frac{\rho (a)^3}{3\epsilon_0 (a)^2}\,\hat{\textbf{j}}-\frac{\rho(a)^3}{3\epsilon_0(3a)^2}\,\hat{\textbf{j}},
\end{equation}

which, after simplification, becomes

\begin{equation}
\label{etotal4}
\vec{E}_{\text{total}}=\frac{\rho (2a)}{3\epsilon_0 }\,\hat{\textbf{j}}-\frac{\rho (a)}{3\epsilon_0 }\,\hat{\textbf{j}}-\frac{\rho(a)}{27\epsilon_0}\,\hat{\textbf{j}}.
\end{equation}

Making the vector sum in equation \eqref{etotal4}, we finally get

\begin{equation}
\label{etotal5}
\vec{E}_{\text{total}}=\frac{8\rho a}{27\epsilon_0}\,\hat{\textbf{j}}.
\end{equation}

Using the numerical values in the SI units (\(a=0.4\,\text{m}\) and \(\rho=2\times 10^{-5}\,\text{C/m}^3\)), we obtain

\begin{equation*}
\vec{E}_{\text{total}}=\frac{8(2\times 10^{-5}\,\text{C/m}^3)(0.4\,\text{m})}{27(8.854\times10^{-12}\,\text{F/m})}\,\hat{\textbf{j}},
\end{equation*}

\begin{equation}
\vec{E}_{\text{total}}\approx 2.68\times 10^{5}\,\text{N/C}\,\hat{\textbf{j}}.
\end{equation}

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