Question

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.$\text{6.90}\times {\text{10}}^{-13}{\text{\hspace{0.17em}C/m}}^{\text{2}}$What is the magnitude of the electron’s acceleration when it reaches radial distances (a)$r=0.500R$and (b)$\mathbf{2}\mathbf{.00}\mathit{R}$?

Short answer

a) The magnitude of the electron’s acceleration when it reaches radial distances$r=0.500\text{R}$ is zero.

b) The magnitude of the electron’s acceleration when it reaches radial distances $r=2.00\text{R}$is $3.43\times {10}^9{\text{m/s}}^2$.

Step-by-step solution

The given data

a) Radius of the metal shell is $R$.

b) The shell is negatively charged.

c) Surface charge density of the shell $\sigma =6.90\times {10}^{-13}{\text{\hspace{0.17em}C/m}}^2$,

Understanding the concept of shell theorems

This problem is based on the two shell theories of electrostatics. That is given as:

a) First shell theorem states that the 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.

b) Second shell theorem states that charged particle inside a shell with charge uniformly distributed on its surface has no net force acting on it due to the shell.

Now using Newton's second law of motion, we can get the required acceleration value that is applied to the body due to the electrostatic force.

Formulae:

The force according to Newton’s second law of motion,$F=ma$ (i)

The electrostatic force due to the two charges in the system, $F=\frac{k{q}_1{q}_2}{r^2}$(ii)

The surface charge density of a sphere, $\sigma =\frac{Q}{4\pi r^2}$ (iii)

a) Calculation of the magnitude of the electron acceleration at r=0.5R.

Here, the radial distance is less than the radius of shell$(r=0.5R<R)$.

Thus, according to second shell theorem of electro statistics, a charged particle inside a shell with charge uniformly distributed on its surface has no net force acting on it. Thus,

the net electrostatic force on the charge is zero. So, using equation (i), we can say that the acceleration value with seize to zero value inside the shell.

Hence, the magnitude of the electron acceleration is zero.

b) Calculation of the magnitude of the electron acceleration at r=2.0R.

As the radial distance of the electron larger than the radius of shell,$(r=2.0R>R)$thus, using the concept of first shell theorem and using the given data and equation (iii) in equation (ii), the magnitude of the electrostatic force can be calculated as follows:

$\begin{array}{c}\left|F\right|=\frac{k\left(4\pi R^2\sigma \right)\left|e\right|}{{\left(2R\right)}^2}(\because q_1=4\pi R^2\sigma ,q_2=e)\\ =k\pi \sigma \left|e\right|\end{array}$

For the given values, the above equation becomes,

$\begin{array}{c}\left|F\right|=(9\times {10}^9{\text{\hspace{0.17em}N.m}}^2{\text{/C}}^2)\pi (6.90\times {10}^{-13}{\text{\hspace{0.17em}C/m}}^2)(1.6\times {10}^{-19}\text{\hspace{0.17em}C})\\ =3.12\times {10}^{-21}\text{\hspace{0.17em}N}\end{array}$

Now, the value of the acceleration of the electron can be given using the above data in equation (i) as follows:

$\begin{array}{c}\left|a_e\right|=\frac{\left|F\right|}{m_e}\\ =\frac{3.12\times {10}^{-21}\text{\hspace{0.17em}N}}{9.1\times {10}^{-31}\text{\hspace{0.17em}kg}}\\ =3.43\times {10}^9{\text{\hspace{0.17em}m/s}}^2\end{array}$

Hence, the value of the acceleration is $3.43\times {10}^9{\text{\hspace{0.17em}m/s}}^2$.