Q19.3-34PE - College Physics (Urone)

Question

(a) What is the potential between two points situated \(10{\rm{ }}cm\) and \(20{\rm{ }}cm\) from a \(3.0{\rm{ }}\mu C\) point charge? (b) To what location should the point at \(20{\rm{ }}cm\) be moved to increase this potential difference by a factor of two?

Step-by-Step Solution

Answer

1. Between the two points potential energy is obtained as: \(\Delta V{\rm{ }} = {\rm{ }}1.3{\rm{ }} \times {\rm{ }}{10^5}{\rm{ }}V\).
2. The potential difference increases by a factor of two is when it is moved towards infinity is said to be the location of the second point.

Step 1: The given data and the required data

Charge of the point charge is:

\(\begin{array}{c}Q{\rm{ }} = {\rm{ }}\left( {3.0{\rm{ }}\mu C} \right)\\ = {\rm{ }}(\frac{{1C}}{{{{10}^6}\mu C}})\\ = {\rm{ }}3.0 \times {10^{ - 6}}{\rm{ }}C\end{array}\)

Distance between the first point and the point charge is:

\(\begin{array}{c}{r_1}{\rm{ }} = {\rm{ }}\left( {10{\rm{ }}cm} \right)\\ = (\frac{{1{\rm{ }}m}}{{100{\rm{ }}cm}})\\ = 0.10{\rm{ }}m\end{array}\)

Distance between the second point and the point charge is:

\(\begin{array}{c}{r_2}{\rm{ }} = {\rm{ }}\left( {20{\rm{ }}cm} \right)\\ = {\rm{ }}(\frac{{1{\rm{ }}m}}{{100{\rm{ }}cm}})\\ = {\rm{ }}0.20{\rm{ }}m\end{array}\)

Determining the potential difference between two points.

Determining the location of the second point such that the potential difference increases by a factor of two.

Step 2: Define Electric Field

The term "electric field" refers to a physical field that surrounds electrically charged particles and exerts force on all other charged particles in the field, either attracting or repelling them.

Step 3: Concepts and Principles

The electron Volt is the energy given to a fundamental charge accelerated through a Electron Volt: It is the energy given to a fundamental charge accelerated through a potential difference of \(\left( {1{\rm{ }}V} \right)\). It is in equation form as:

\(\begin{array}{c}1{\rm{ }}eV{\rm{ }} = {\rm{ }}\left( {1.60{\rm{ }}x{\rm{ }}{{10}^{ - 19}}{\rm{ }}C} \right)\left( {1{\rm{ }}V} \right)\\ = {\rm{ }}1.60{\rm{ }}x{\rm{ }}{10^{ - 19}}{\rm{ }}J\end{array}\)…………………….(I)

The electric potential energy is associated with a pair of point charges separated by the distance \(r\) is in the form of:

\({U_e} = k\frac{{{q_1}{q_2}}}{r}\)………………..(II)

Here, the value  is the Coulomb's constant, a proportionality constant. The joule/coulomb is referred to as the unit of electric potential and is referred to as the volt \((V)\).

Step 4: Evaluating potential difference between two points

1. Electric potential at the location of the first point is found from the above equation as:

\({V_1}{\rm{ }} = {\rm{ }}\frac{{kQ}}{{{r_1}}}\)

Entering the values and we get is:

 

The electric potential at the second point is:

\(\begin{array}{c}{V_2}{\rm{ }} = {\rm{ }}\frac{{(8.99 \times {{10}^9}{\rm{ }}N.{m^2}/C){\rm{ }}(3.0 \times {{10}^{ - 6}}{\rm{ }}C)}}{{0.20{\rm{ }}m}}\\ = {\rm{ }}1.3485 \times {10^5}{\rm{ }}V\end{array}\)

Difference between the two points is then determined as:

\(\begin{array}{c}\Delta V{\rm{ }} = {\rm{ }}{V_1}{\rm{ }} - {\rm{ }}{V_2}\\ = {\rm{ }}2.697{\rm{ }} \times {\rm{ }}{10^5}{\rm{ }}V{\rm{ }} - {\rm{ }}1.3485{\rm{ }} \times {\rm{ }}{10^5}{\rm{ }}V\\ = {\rm{ }}1.3 \times {10^5}{\rm{ }}V{\rm{ }}(1.3485{\rm{ }} \times {\rm{ }}{10^5}{\rm{ }}V)\end{array}\)

Step 5: Evaluating difference of two points two times

1. The potential difference between the two points to be two times \(\Delta V\), we obtain is:

\(\begin{array}{c}\Delta {V'} = {\rm{ }}2{\rm{ }}(\Delta {V'})\\ = {\rm{ }}{V_1}{\rm{ }} - {\rm{ }}{V_2}\\ = 2{\rm{ }}(1.3485{\rm{ }} \times {\rm{ }}{10^5}{\rm{ }}V)\\ = {\rm{ }}{V_1}{\rm{ }} - {\rm{ }}{V_2}\end{array}\)

Substituting the values and we get:

\(\begin{array}{c}2.697 \times {10^5}\;V{\rm{ }} = {\rm{ }}\frac{{kQ}}{{{r_1}}}{\rm{ }} - {\rm{ }}\frac{{kQ}}{{{r_2}}}\\\frac{{kQ}}{{{r_2}}}{\rm{ }} = {\rm{ }}\frac{{kQ}}{{{r_1}}}{\rm{ }} - {\rm{ }}2.697 \times {10^5}\;V\\{r_2}{\rm{ }} = {\rm{ }}\frac{{kQ}}{{\frac{{kQ}}{{{r_1}}}{\rm{ }} - {\rm{ }}2.697{\rm{ }} \times {\rm{ }}{{10}^5}\;V}}\\{r_2}{\rm{ }} = {\rm{ }}\frac{{kQ}}{{2.697{\rm{ }} \times {\rm{ }}{{10}^5}\;V{\rm{ }} - {\rm{ }}2.697{\rm{ }} \times {\rm{ }}{{10}^5}\;V}}\\{r_2}{\rm{ }} = {\rm{ }}\frac{{kQ}}{0}\\ = {\rm{ }}\infty \end{array}\)

Therefore, we get:

1. Potential difference between the two points is: \(\Delta V{\rm{ }} = {\rm{ }}1.3{\rm{ }} \times {\rm{ }}{10^5}{\rm{ }}V\).
2. Location of the second point so that the potential difference increases by a factor of two is when it move towards infinity.