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?

Short 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-by-step solution

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.

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.

Concepts and Principles

Theelectron Volt is the energy given to a fundamental charge accelerated through aElectron 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 \((V)\).

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:

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

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.