Question

How far from a \(1.00{\rm{ }}\mu C\) point charge will the potential be \(100{\rm{ }}V\)? At what distance will it be \(2.00 \times {10^2}{\rm{ }}V\)?

Short answer

(a) At a distance of\(r = 90.0{\rm{ }}m\)a\(1.00{\rm{ }}\mu C\)point charge will the potential be\(100{\rm{ }}V\).

(b) At a distance of \(r' = 45.0\;m\) a \(1.00{\rm{ }}\mu C\) point charge will the potential be \(2.00 \times {10^2}{\rm{ }}V\).

Step-by-step solution

Given Information

• The potential charge value –\(1.00{\rm{ }}\mu C\)
• The given potential –\(100{\rm{ }}V\)
• The given potential – \(2.00 \times {10^2}{\rm{ }}V\)

Concept Introduction

Electric Potential Due to a Single Charge: Place a test charge\(q\)at the desired site, and then calculate the electric potential energy\({U_{Qq}}\)of the system containing the test charge and the source charge that generates the field to determine the electric potential owing to a single charge at that location.

The ratio -\(V = \frac{{{U_{Qq}}}}{q} = \frac{{kQ}}{r}...(1)\)determines the electric potential there, where\(k = 8.99 \times {10^9}{\rm{ }}N \times {m^2}/C\)is a proportionality constant known as the Coulomb's constant. The joule/coulomb\(\left( {J/C} \right)\), often known as the volt\(\left( V \right)\)is the unit of electric potential.

Calculation for the distance

(a)

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

\(V = \frac{{kq}}{r}\)

Solving for\(r\), it is obtained –

\(r = \frac{{kq}}{V}\)

Entering the values for\(k,q\), and \(V\)–

\(\begin{array}{c}r = \frac{{\left( {8.99 \times {{10}^9}\;N \times {m^2}/C} \right)\left( {1.00 \times {{10}^{ - 6}}{\rm{ }}C} \right)}}{{100\;V}}\\ = 90.0\;m\end{array}\)

Calculation for the distance

(b)

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

\(V' = \frac{{kq}}{{r'}}\)

Solving for\(r'\), it is obtained –

\(r' = \frac{{kq}}{{V'}}\)

Entering the values for\(k,q\), and \(V'\)–

\(\begin{array}{c}r' = \frac{{\left( {8.99 \times {{10}^9}\;N \times {m^2}/C} \right)\left( {1.00 \times {{10}^{ - 6}}{\rm{ }}C} \right)}}{{2.00 \times {{10}^2}\;V}}\\ = 45.0\;m\end{array}\)

Therefore, the distances are obtained as \(r = 90.0\;m\)and \(r' = 45.0\;m\).