P20B2015 – Overlapping Spheres

The insulating sphere has a radius $R$ and we have no reason to assume that the electric field inside the insulating material will be the same as outside of the sphere. Thus, we differentiate two regions: inside the insulating sphere $\left(r\le R\right)$ and outside the insulating sphere $\left(r>R\right)$.

Electric field inside the sphere $\left(r\le R\right)$:
The electric flux through our Gaussian surface simplifies because everywhere on the surface of the Gaussian sphere the electric field is along $d\overrightarrow{A}$ and has constant magnitude

\begin{aligned}
\oiint_A{\overrightarrow{E}\cdot d\overrightarrow{A}}&=\iint_A{E{\mathrm{cos} \left(0\right)\ }dA} \\
\\
&=\iint_A{E{\mathrm{cos} \left(0\right)\ }dA} \\
\\
&=E{\mathrm{cos} \left(0\right)\ }\iint_A{dA} \\
\\
&=E\cdot 4\pi r^2
\end{aligned}

where the area of the Gaussian sphere is equal to $A=4\pi r^2$.

The Gaussian sphere encloses a sphere of charge with volume $V=4\pi r^3/3$ and therefore the amount of enclosed charge is equal to

Q_{enc}=\rho V=\rho \cdot \frac{4\pi r^3}{3}

By Gauss’s Law, we conclude that the magnitude of the electric field created inside the sphere of charge is given by

E\cdot 4\pi r^2=\rho \cdot \frac{4\pi r^3}{3{\varepsilon }_0}\ \ \ \ \Rightarrow \ \ \ \ \ E=\frac{\rho r}{3{\varepsilon }_0}

Thus, the electric field created inside the insulating sphere is equal to

\boxed{\overrightarrow{E}=\frac{\rho r}{3{\varepsilon }_0}\ \hat{r}}

where $\rho >0$ yields radially outward electric field lines and $\rho < 0$ yields radially inward electric field lines.

Outside the sphere $\left(r>R\right)$:
The electric flux through our Gaussian surface simplifies because everywhere on the surface of the Gaussian sphere the electric field is along $d\overrightarrow{A}$ and has constant magnitude

\begin{aligned}
\oiint_A{\overrightarrow{E}\cdot d\overrightarrow{A}}&=\iint_A{E{\mathrm{cos} \left(0\right)\ }dA} \\
\\
&=\iint_A{E{\mathrm{cos} \left(0\right)\ }dA} \\
\\
&=E{\mathrm{cos} \left(0\right)\ }\iint_A{dA} \\
\\
&=E\cdot 4\pi r^2
\end{aligned}

where the area of the Gaussian sphere is equal to $A=4\pi r^2$.

The Gaussian sphere encloses a sphere of charge with volume $V=4\pi R^3/3$ and therefore the amount of enclosed charge is equal to

Q_{enc}=\rho V=\rho \cdot \frac{4\pi R^3}{3}

By Gauss’s Law, we conclude that the magnitude of the electric field created outside the sphere of charge is given by

E\cdot 4\pi r^2=\rho \cdot \frac{4\pi R^3}{3{\varepsilon }_0}\ \ \ \ \Rightarrow \ \ \ \ \ E=\frac{\rho R^3}{3{\varepsilon }_0r^2}

Thus, the electric field created outside the insulating sphere is equal to

\boxed{\overrightarrow{E}=\frac{\rho R^3}{3{\varepsilon }_0r^2}\ \hat{r}}

where $\rho >0$ yields radially outward electric field lines and $\rho < 0$ yields radially inward electric field lines.

The graph of the electric field $\left(\rho >0\right)$ against the radial distance $r$ is drawn below

Note that this volume corresponds to the overlapping region if we were to consider two full spheres interpenetrating each other. The two centers $O$ and $A$ are separated by a distance $d$. Determine the electric field created by this charge distribution at any point $M$ located in the overlapping region.

We let ${\hat{r}}_+$ and ${\hat{r}}_-$ denote the radially outward unit vectors centered on spheres with charge density $\rho $ and $-\rho $ respectively. We then draw the electric fields ${\overrightarrow{E}}_+$ and ${\overrightarrow{E}}_-$ created at point $P$ by the positive sphere and the negative sphere respectively.

The net electric field at point $P$ by superposition is therefore equal to

\overrightarrow{E}=\frac{\rho r_+}{3{\varepsilon }0}\ \ {\hat{r}}_++\frac{\rho r_-}{3{\varepsilon }0}\ \ {\hat{r}}_-

where $r_+$ denotes the distance between the center of the positive sphere and $P$ while $r_-$ denotes the distance between the center of the negative sphere and $P$.

Noting that $r_+\ {\hat{r}}_+ + r-\ \ {\hat{r}}_-=\overrightarrow{d}$ as shown in the figure below

we conclude that the electric field in the overlapping region is uniform and expression in terms of $\overrightarrow{d}$ as follows

\boxed{\overrightarrow{E}=\frac{\rho r_+}{3{\varepsilon }0}\ \ {\hat{r}}_++\frac{\rho r_-}{3{\varepsilon }0}\ \ {\hat{r}}_-=\frac{\rho }{3{\varepsilon }0}\ \ \left(r_+\ \ {\hat{r}}_++r_-\ \ {\hat{r}}_-\right)=\frac{\rho \overrightarrow{d}}{3{\varepsilon }_0}}

The field within the overlapping spheres is uniform and the field lines are horizontal as shown