Chapter 33, Problem 43A

In a hydrogen atom, the proton and the electron $\left(q=1.6\times {10}^{-19}C\right)$ are separated by an average distance of $5\times {10}^{-11}m$.

a. Calculate the force that the proton exerts on the electron at this distance.

b. Show that the electric field strength at the average location of the electron is an enormous $6\times {10}^{11}N/C$.

To determine

The force exerted by proton on electron.

Answer to Problem 43A

The force = $9.216\times {10}^{-8}\text{N}$ .

Explanation of Solution

Given:

Charge on electron and proton, $q=1.6\times {10}^{-19}C.$

Distance between proton and electron, $r=5\times {10}^{-11}\text{ m}\text{.}$

Formula used:

The force exerted is given as,

  $F=\frac{k{q}^2}{r^2}$

Here, k is the Coulomb’s constant.

Calculation:

The force exerted by proton on electron is:

  $\begin{array}{l}F=\frac{k{q}^2}{r^2}\\ F=\frac{9\times {10}^9\times {\left(1.6\times {10}^{-19}\right)}^2}{{\left(5\times {10}^{-11}\right)}^2}\\ F=9.216\times {10}^{-8}\text{ N}\text{.}\end{array}$

Conclusion:

The force exerted by proton on electron is $9.216\times {10}^{-8}\text{N}$ .

To determine

The electric field at average location of electron.

Answer to Problem 43A

The electric field = 6×10¹¹ N/C.

Explanation of Solution

Given data:

Charge on electron and proton, q = 1.6×10-19 C.

Distance between proton and electron, r = 5×10-11 m.

Formula used:

The electric field is given as,

  $E=\frac{kq}{r^2}$

Here, k is the Coulomb’s constant.

Calculation:

The electric field at the average position of an electron:

  $\begin{array}{l}E=\frac{kq}{r^2}\\ E=\frac{9\times {10}^9\times 1.6\times {10}^{-19}}{{\left(5\times {10}^{-11}\right)}^2}\\ E=5.76\times {10}^{11}\text{ N/C}\text{.}\\ E\approx 6\times {10}^{11}\text{ N/C}\text{.}\end{array}$

Conclusion:

The electric field = 6×10¹¹ N/C.