Q25PE - College Physics (Urone)

Question

What is the potential $\mathbf{0}\mathbf{.}\mathbf{530}\mathbf{\times }{\mathbf{10}}^{\mathbf{-}\mathbf{10}}\mathbf{ }\mathbf{m}$ from a proton (the average distance between the proton and electron in a hydrogen atom)?

Step-by-Step Solution

Answer

The potential at $0.530\times {10}^{-10} \mathrm{m}$ from a proton is $\mathrm{V}=27.2 \mathrm{V}$.

Step 1: Given Data

Distance from proton is $0.530\times {10}^{-10} \mathrm{m}$

Step 2: Concept Introduction

Electric Potential Due to a Single Charge: To determine the electric potential V due to a single source charge Q at a specific location, place a test charge q at that location and determine the electric potential energy U_Qq of a system containing the test charge and the source charge that creates the field. The electric potential is given as–

$\mathbf{V}\mathbf{=}\frac{{\mathbf{U}}_{\mathbf{Qq}}}{\mathbf{q}}\mathbf{=}\frac{\mathbf{kQ}}{\mathbf{r}}\mathbf{.}\mathbf{.}\mathbf{.}\mathbf{.}\mathbf{.}\mathbf{.}\mathbf{.}\mathbf{.}\mathbf{.}\mathbf{.}\mathbf{.}\mathbf{.}\mathbf{.}\mathbf{.}\mathbf{.}\mathbf{.}\mathbf{.}\mathbf{.}\mathbf{.}\mathbf{.}\mathbf{.}\mathbf{.}\mathbf{.}\mathbf{.}\mathbf{.}\mathbf{.}\mathbf{.}\mathbf{\left(}\mathbf{1}\mathbf{\right)}$

Where $\mathbf{k}\mathbf{=}\mathbf{8}\mathbf{.}\mathbf{99}\mathbf{\times }{\mathbf{10}}^{\mathbf{9}}\mathbf{ }\mathbf{N}\mathbf{·}{\mathbf{m}}^{\mathbf{2}}\mathbf{/}\mathbf{C}$ is a proportionality constant called Coulomb's constant.

Step 3: Calculation for Electric Potential

The electric potential a distance r from the proton is found from Equation (1) –

$\mathbf{V}\mathbf{=}\frac{\mathbf{kq}}{\mathbf{r}}$

Entering the values for k, q, and r–

$\begin{array}{rcl}\mathbf{V}& \mathbf{=}& \frac{\left(8.99\times {10}^9 N\times m^2/C\right)\left(1.60\times {10}^{-19}C\right)}{\mathbf{0}\mathbf{.}\mathbf{530}\mathbf{\times }{\mathbf{10}}^{\mathbf{-}\mathbf{10}}\mathbf{ }\mathbf{m}}\\ & \mathbf{=}& \mathbf{27}\mathbf{.}\mathbf{2}\mathbf{ }\mathbf{V}\\ & & \end{array}$

Therefore, the potential is obtained as $\mathbf{V}\mathbf{=}\mathbf{27}\mathbf{.}\mathbf{2}\mathbf{ }\mathbf{V}$.