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Found in: Page 697

### College Physics (Urone)

Book edition 1st Edition
Author(s) Paul Peter Urone
Pages 1272 pages
ISBN 9781938168000

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

The potential at ${0}{.}{530}{×}{{10}}^{-10}{}{\mathrm{m}}$ from a proton is ${\mathrm{V}}{=}{27}{.}{2}{}{\mathrm{V}}$.

See the step by step solution

## Step 1: Given Data

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

## 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 UQq 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 ${\mathbf{k}}{\mathbf{=}}{\mathbf{8}}{\mathbf{.}}{\mathbf{99}}{\mathbf{×}}{{\mathbf{10}}}^{{\mathbf{9}}}{\mathbf{}}{\mathbf{N}}{\mathbf{·}}{{\mathbf{m}}}^{{\mathbf{2}}}{\mathbf{/}}{\mathbf{C}}$ 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–

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

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