Chapter 21, Problem 67A

(a)

To determine

The direction and magnitude of the electric field.

Answer to Problem 67A

  $E=1.17\times {10}^{13}\text{ N/C}$ , outward

Explanation of Solution

Given:

Charge of lead is $\text{82}$ protons.

The field should be measure at distance $d=1.0\times {10}^{-10}\text{ m }$ from the nucleus

Formula used:

Total charge on the lead atom is,

  $\text{Q = }\left(\text{82 protons}\right)\left(\text{charge of one proton}\right)$ $\left(1\right)$

Charge of one proton is $1.6\times {10}^{-19}\text{ C}$

If $F$ is the electrostatic force on the test charge $q$ at some point, the electric field $\left(E\right)$ on that charge at the same point is given by,

  $E=\frac{F}{q}$ $\left(2\right)$

Iftwo charges $q$ and $Q$ are separated by a distance $d$ , then Coulomb’s law can be written as,

  $F\text{ = K}\frac{qQ}{d^2}$

Where $\text{ K}$ is a constant, $\text{ K}=9\times {10}^9\text{ N}{\text{.m}}^2/C^2$

Substituting for $F$ in equation $\left(2\right)$ ,

  $E=\frac{\left(\text{K}\frac{qQ}{d^2}\right)}{q}=\text{K}\frac{Q}{d^2}$ $\left(3\right)$

Calculation:

Substituting the numerical values in equation $\left(1\right)$ ,

  $\text{Q = }\left(\text{82}\right)\left(1.6\times {10}^{-19}\text{ C}\right)=1.31\times {10}^{-17}\text{ C}$

Now substitute the numerical values in equation $\left(3\right)$ ,

  $E=\frac{\left(9\times {10}^9\text{ N}{\text{.m}}^2/{\text{C}}^2\right)\left(1.31\times {10}^{-17}\text{ C}\right)}{{\left(1.0\times {10}^{-10}\text{ m}\right)}^2}$

  $=\frac{11.79\times {10}^{-8}\text{ N}{\text{.m}}^2/\text{C}}{1.0\times {10}^{-20}{\text{ m}}^2}$

  $=1.17\times {10}^{13}\text{ N/C}$

Conclusion:

The electric field at $1.0\times {10}^{-10}\text{ m }$ from the nucleus is $1.17\times {10}^{13}\text{ N/C}$ acting outward from the nucleus.

(b)

To determine

The magnitude and direction of force exerted on an electron located at a given distance from the nucleus.

Answer to Problem 67A

  $F=\text{1}\text{.8}\times {10}^{-6}\text{ N}$ , directed towards the nucleus.

Explanation of Solution

Given:

Charge of lead is $\text{82}$ protons.

The electron is located at a distance $d=1.0\times {10}^{-10}\text{ m }$ from the nucleus

Formula used:

If $F$ is the electrostatic force on the test charge $q$ at some point, the electric field $\left(E\right)$ on that charge at the same point is given by,

  $E=\frac{F}{q}$

  $\Rightarrow F=Eq$ $\left(4\right)$

Where, $E=1.17\times {10}^{13}\text{ N/C}$ from the part (a)

$q$ is the charge of an electron, $q=-1.6\times {10}^{-19}\text{ C}$

Calculation:

Substituting the numerical values in equation $\left(4\right)$ ,

  $F=\left(1.17\times {10}^{13}\text{ N/C}\right)\left(-1.6\times {10}^{-19}\text{ C}\right)$

  $=-\text{ 1}\text{.8}\times {10}^{-6}\text{ N}$ , toward the nucleus

Conclusion:

The force exerted on an electron located at $1.0\times {10}^{-10}\text{ m }$ from the nucleus is $\text{1}\text{.8}\times {10}^{-6}\text{ N}$ and is directed towards the nucleus.