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Q25PE

Expert-verifiedFound in: Page 697

Book edition
1st Edition

Author(s)
Paul Peter Urone

Pages
1272 pages

ISBN
9781938168000

**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)?**

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

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

**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}**

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

**Where**** ${\mathbf{k}}{\mathbf{=}}{\mathbf{8}}{\mathbf{.}}{\mathbf{99}}{\mathbf{\times}}{{\mathbf{10}}}^{{\mathbf{9}}}{\mathbf{}}{\mathbf{N}}{\mathbf{\xb7}}{{\mathbf{m}}}^{{\mathbf{2}}}{\mathbf{/}}{\mathbf{C}}$**** is a proportionality constant called Coulomb's constant. **

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 ${\mathbf{V}}{\mathbf{=}}{\mathbf{27}}{\mathbf{.}}{\mathbf{2}}{\mathbf{}}{\mathbf{V}}$.

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