Q19.3-31PE - College Physics (Urone)

Question

A research Van de Graaff generator has a \(2.00 - m\)-diameter metal sphere with a charge of \(5.00{\rm{ }}mC\) on it. (a) What is the potential near its surface? (b) At what distance from its centre is the potential \(1.00{\rm{ }}MV\)? (c) An oxygen atom with three missing electrons is released near the Van de Graaff generator. What is its energy in \(MeV\) at this distance?

Step-by-Step Solution

Answer

1. Near the generator's surface the electric potential \(V\) is obtained as: \(V{\rm{ }} = {\rm{ }}45.0{\rm{ }} \times {\rm{ }}{10^6}{\rm{ }}V\).
2. From the centre of the Van de Graaff generator the electric potential of the distance \(r'\)  is obtained as: \(r' = {\rm{ }}45.0{\rm{ }}m\).
3. From the centre of the Van de Graaff generator the electric potential energy in \(MeV\) of an oxygen atom with missing three electron located at a distance \(r'\)  is obtained as: \({U_e}{\rm{ }} = {\rm{ }}132{\rm{ }}MeV\).

Step 1: The given data

Diameter of Van de Graaff generator is:

\(D{\rm{ }} = {\rm{ }}2.00{\rm{ }}m\).

Radius of the generator Van de Graaff generator is:

\(\begin{array}{c}r{\rm{ }} = {\rm{ }}D/2{\rm{ }}\\ = {\rm{ }}1.00{\rm{ }}m\end{array}\).

Charge on Van de Graaff generator is: 

\(\begin{array}{c}Q = (5.00mC)\\ = (\frac{{1C}}{{1000mC}})\\ = 0.00500C\end{array}\)

Electric potential a distance \(r'\) from centre of Van de Graaff generator is: 

\(\begin{array}{c}{V'} = (1.00mV)\\ = (\frac{{{{10}^6}V}}{{1V}})\\ = 1.00 \times {10^6}V\end{array}\)

Oxygen atom in the part (c) is missing three electrons.

Charge of an electron is:

\(\begin{array}{c}q{\rm{ }} = {\rm{ }} - e{\rm{ }}\\ = {\rm{ }} - 1.60{\rm{ }}x{\rm{ }}{10^{ - 19}}{\rm{ }}C\end{array}\)

Step 2: Define Electric Field

A physical field that surrounds electrically charged particles and exerts force on all other charged particles in the field, either attracting or repelling them, is referred to as an electric field.

Step 3: Concepts and Principles

A single charge's electric potential is as follows: to ascertain the electric potential caused by a single charge source at a particular place. The electric potential energy \({U_{Qq}}\) of a system including the test charge and the source charge that generates the field is measured after placing a test charge there. The electric potential there is then equal to the ratio given by:

\(\begin{array}{c}V = \frac{{{U_{Qq}}}}{q}\\ = \frac{{kQ}}{r}\end{array}\)…………….(I)

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 \((V)\).

Electric Potential Energy: If a charged particle \(q\) is positioned at a location where the electric potential of a charged object \(V\) is claimed to be, the particle-object system's electric potential energy is:

\(Ue{\rm{ }} = {\rm{ }}qV\)…………...(II)

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:

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

Step 4: The required data

In part (a), determining the electric potential \(V\) near the generator's surface

In part (b), determining the distance \(r'\) from the centre of the Van de Graaff generator and the electric potential is \(V'{\rm{ }} = {\rm{ }}1.00{\rm{ }}MV\).

In part (c), determining the electric potential energy in \(MeV\) of an oxygen atom with missing three electron located a distance \(r'\) and from the centre of the Van de Graaff generator.

Step 5: Evaluating electric potential near generator’s surface

1. The electric potential near the Van de Graaff generator's surface is evaluated from the first equation as:

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

Entering the values and then we obtain:

\(\begin{array}{c}V = \frac{{(8.99 \times {{10}^9}N.{m^2}/C)(5.00 \times {{10}^{ - 3}}C)}}{{1.00m}}\\ = 45.0 \times {10^6}V\end{array}\)

Step 6: Evaluating the electric potential a distance from the centre of the Van de Graaff generator

1. The electric potential a distance \(r'\) from the centre of the Van de Graaff generator is evaluated from the first equation as:

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

Solving for the value of \(r'\), we get:

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

Substituting the numerical values as:

\(\begin{array}{c}r' = \frac{{(8.99 \times {{10}^9}N.{m^2}/C)(5.00 \times {{10}^{ - 3}}C)}}{{1.00 \times {{10}^6}V}}\\ = 45.0m\end{array}\)

Step 7: Evaluating electric potential energy of the oxygen atom at a distance from the centre of the Van de Graaff generator

1. Charge of the oxygen atom with the three missing electrons is:

\(\begin{array}{c}Q =  + 3e\\ = 3(1.60 \times 10{}^{ - 19}C)\\ = 4.80 \times 10{}^{ - 19}C\end{array}\)

The electric potential energy of the oxygen atom a distance \(r'\) from the centre of the Van de Graaff generator is evaluated from the second equation as:

\(\begin{array}{c}{U_e} = q\Delta V\\ = q\left( {V - {V'}} \right)\\ = \left( {4.80 \times {{10}^{ - 19}}C} \right)\left( {45.0 \times {{10}^6}\;V - 1.00 \times {{10}^6}\;V} \right)\\ = 2.112 \times {10^{ - 11}}\;J\end{array}\)

Converting this energy into MeV using the conversion factor in the third equation as:

\(\begin{array}{c}{U_e} = \left( {2.112 \times {{10}^{ - 11}}\;J} \right)\left( {\frac{{1eV}}{{1.60 \times {{10}^{ - 19}}\;J}}} \right)\left( {\frac{{1MeV}}{{{{10}^6}eV}}} \right)\\ = 132MeV\end{array}\)

Therefore, we get:

1. Electric potential \(V\) near the generator's surface Is: \(V{\rm{ }} = {\rm{ }}45.0{\rm{ }} \times {\rm{ }}{10^6}{\rm{ }}V\).
2. Distance \(r'\) from the centre of the Van de Graaff generator the electric potential is: \(r' = {\rm{ }}45.0{\rm{ }}m\).
3. Electric potential energy in \(MeV\) of an oxygen atom with missing three electron located a distance \(r'\) from the centre of the Van de Graaff generator is: \({U_e}{\rm{ }} = {\rm{ }}132{\rm{ }}MeV\).