P20B2015 – Force From a Uniformly Charged Sphere

Before we can find the force exerted by the sphere on the rod, we must first determine the electric field created by the uniformly charged solid sphere outside of the sphere $\left(r > R\right)$. Given the symmetry of this charge distribution, we apply Gauss’s Law to derive the electric field created by the solid sphere and choose a Gaussian sphere of radius $r>R$ centered on the solid sphere as our Gaussian surface.

Electric field created by the solid sphere (outside $r > R$):

To derive the electric field outside of the sphere, we choose a Gaussian sphere of radius $r>R$ centered on the insulating sphere with an outward oriented normal as shown in the figure below.

The electric field is constant in magnitude and direction on the Gaussian surface and the electric flux through the Gaussian sphere is therefore equal to

\oiint_A{{\overrightarrow{E}}_S\cdot d\overrightarrow{A}}=\oiint_A{E_S{\mathrm{cos} \left(0\right)\ }dA}=E_S\cdot 4\pi r^2

The charge enclosed by the Gaussian sphere is equal to the volume charge density $\rho $ multiplied by the volume of the solid sphere i.e.

Q_{enc}=\rho \left(\frac{4}{3}\pi R^3\right)=\frac{4\rho }{3}\pi R^3

Thus, by Gauss’s Law, the electric field outside the sphere is given by

E_S\cdot 4\pi r^2=\frac{4\rho }{3{\varepsilon }_0}\pi R^3\ \ \ \ \Rightarrow \ \ \ \ \ \boxed{\overrightarrow{E_S}=\frac{\rho R^3}{3{\varepsilon }_0r^2}\hat{r}}

Force acting on the line of charge:

We let $\lambda =Q/d$ denote the uniform linear charge density of the rod and, in order to derive the force acting on the rod from the sphere, we consider an infinitesimal fraction $dx$ with charge $dq=\lambda dx$ of the rod and write the force $dF_E$ acting on it.

The infinitesimal force $d\overrightarrow{F_E}$ acting on the rod due to the electric field ${\overrightarrow{E}}_S$ is equal to

d\overrightarrow{F_E}=dq\ \overrightarrow{E_S}=\frac{\rho R^3}{3{\varepsilon }_0x^2}\cdot \frac{Q}{d}dx\ \hat{x}

Integrating over the entire line of charge, we derive the total force from the sphere

\begin{aligned}
\overrightarrow{F_E}&=\int^{x=R+d}_{x=R}{\frac{\rho QR^3}{3{\varepsilon }_0d}\cdot \frac{dx}{x^2}}\ \ \hat{x} \\
\\
&=\frac{\rho QR^3}{3{\varepsilon }_0d}{\left[-\frac{1}{x}\right]}^{R+d}_R\ \hat{x} \\
\\
&=\frac{\rho QR^3}{3{\varepsilon }_0d}\left[\frac{1}{R}-\frac{1}{R+d}\right]\ \ \hat{x} \\
\\
&=\frac{\rho QR}{3{\varepsilon }_0(R+d)}\ \hat{x}
\end{aligned}

The force from the sphere on the uniform line charge with total charge $Q$ is equal to

\boxed{\overrightarrow{F_E}=\frac{\rho QR}{3{\varepsilon }_0(R+d)}\ \hat{x}}