Question 21.71: In a spherical metal shell of radius R, an electron is shot ......
In a spherical metal shell of radius R, an electron is shot from the center directly toward a tiny hole in the shell, through which it escapes. The shell is negatively charged with a surface charge density (charge per unit area) of 6.90 × 10^{-13} C/m².What is the magnitude of the electron’s acceleration when it reaches radial distances (a) r = 0.500R and (b) 2.00R?
(a) The second shell theorem states that a charged particle inside a shell with charge uniformly distributed on its surface has no net force acting on it due to the shell. Thus, inside the spherical metal shell at r = 0.500R < R, the net force on the electron is zero, and therefore, a = 0.
(b) The first shell theorem states that a charged particle outside a shell with charge uniformly distributed on its surface is attracted or repelled as if the shell’s charge were concentrated as a particle at its center. Thus, the magnitude of the Coulomb force on the electron at r = 2.00R is
\begin{aligned} F & =k \frac{Q|e|}{r^2}=k \frac{\left(4 \pi R^2 \sigma\right)|e|}{(2.0 R)^2}=k \pi \sigma|e| \\ & =\left(8.99 \times 10^9 \,N \cdot m ^2 / C ^2\right) \pi\left(6.90 \times 10^{-13} \,C / m ^2\right)\left(1.60 \times 10^{-19} \,C \right) \\ & =3.12 \times 10^{-21} \,N , \end{aligned}
and the corresponding acceleration is
a=\frac{F}{m}=\frac{3.12 \times 10^{-21} \,N }{9.11 \times 10^{-31} \,kg }=3.43 \times 10^9 \,m / s ^2 .