Chapter 17, Problem 18P

(a)

To determine

The electric potential.

Answer to Problem 18P

The electric potential is $5.76\times {10}^5\text{V}$ .

Explanation of Solution

Given info:

The distance is $r=2.5\times {10}^{-15}\text{m}$ .

The charge on electron is $q=1.6\times {10}^{-19}\text{C}$ .

Formula used:

The electrical potential at a point due to a charge $q$ is,

  $V=\frac{kQ}{r}$

The Coulomb’s constant is $k$ and its value is $9\times {10}^9\text{N}\cdot {\text{m}}^{\text{2}}{\text{/C}}^2$ .

Calculation:

Substituting the given values,

  $\begin{array}{c}V=\frac{\left(9\times {10}^9{\text{Nm}}^{\text{2}}/{\text{C}}^{\text{2}}\right)\left(1.6\times {10}^{-19}\text{C}\right)}{2.5\times {10}^{-15}\text{m}}\\ =5.76\times {10}^5\text{V}\end{array}$

Conclusion:

Thus, the electric potential is $5.76\times {10}^5\text{V}$ .

(b)

To determine

The electric potential energy of the system that consists of two protons is $9.22\times {10}^{-14}\text{J}$ .

Answer to Problem 18P

The electric potential energy of the system is $9.22\times {10}^{-14}\text{J}$ .

Explanation of Solution

Given info:

The distance is $2.5\times {10}^{-15}\text{m}$ .

The charge on proton is $q=1.6\times {10}^{-19}\text{C}$ .

Formula used:

The electrical potential due to two protons is,

  $U=\frac{k{Q}_1{Q}_2}{r}$

The Coulomb’s constant is $k$ and its value is $9\times {10}^9\text{N}\cdot {\text{m}}^{\text{2}}{\text{/C}}^2$ .

Calculation:

Substituting the given values,

  $\begin{array}{c}U=\frac{\left(9\times {10}^9{\text{Nm}}^{\text{2}}/{\text{C}}^{\text{2}}\right){\left(1.6\times {10}^{-19}\text{C}\right)}^2}{2.5\times {10}^{-15}\text{m}}\\ =9.22\times {10}^{-14}\text{J}\end{array}$

Conclusion:

Thus, the electric potential energy of the system that consists of two protons is $9.22\times {10}^{-14}\text{J}$ .