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Three charges, \(q_{1}, q_{2}\), and \(q_{3}\), are located at the corners of an equilateral triangle with side length of \(1.20 \mathrm{~m}\). Find the work done in each of the following cases: a) to bring the first particle, \(q_{1}=1.00 \mathrm{pC},\) to \(P\) from infinity b) to bring the second particle, \(q_{2}=2.00 \mathrm{pC}\) to \(Q\) from infinity c) to bring the last particle, \(q_{3}=3.00 \mathrm{pC}\) to \(R\) from infinity d) Find the total potential energy stored in the final configuration of \(q_{1}\) \(q_{2},\) and \(q_{3}\).

Short Answer

Expert verified
Answer: The work done to bring \(q_{1}\) to \(P\) is 0 J, \(q_{2}\) to \(Q\) is 1.499 x \(10^{-11}\) J, and \(q_{3}\) to \(R\) is 4.498 x \(10^{-11}\) J. The total potential energy stored in the final configuration is 5.997 x \(10^{-11}\) J.

Step by step solution

01

Calculate work done for each charge

a) Work done to bring \(q_{1}\) to \(P\) from infinity: As there are no charges initially at \(P\), \(Q\), or \(R\), there is no force acting on \(q_{1}\), so the work done to bring \(q_{1}\) to \(P\) is 0. b) Work done to bring \(q_{2}\) to \(Q\) from infinity: The only force acting on \(q_{2}\) is from \(q_{1}\). The distance between \(P\) and \(Q\) (\(r_{1,2}\)) is 1.2 m. The work done can be calculated using the potential energy formula: \(W_{2} = U_{1,2} = k \cdot \frac{q_{1}q_{2}}{r_{1,2}}\) c) Work done to bring \(q_{3}\) to \(R\) from infinity: The forces acting on \(q_{3}\) are from both \(q_{1}\) and \(q_{2}\). The distance between points \(P\) and \(R\) (\(r_{1,3}\)) and \(Q\) and \(R\) (\(r_{2,3}\)) is also 1.2 m. The work done is given by: \(W_{3} = U_{1,3} + U_{2,3} = k \cdot (\frac{q_{1}q_{3}}{r_{1,3}} + \frac{q_{2}q_{3}}{r_{2,3}})\)
02

Calculate the total potential energy stored

The total potential energy stored in the final configuration is the sum of potential energies of each pair of charges: \(U_{total} = U_{1,2} + U_{1,3} + U_{2,3}\) Now, we can calculate the numerical values for each work done and the total potential energy stored.
03

Calculate numerical values

\(W_{2} = k \cdot \frac{q_{1}q_{2}}{r_{1,2}} = (8.99 \times 10^{9} \mathrm{Nm^2/C^2}) \cdot \frac{(1.00 \times 10^{-12}\mathrm{C})(2.00 \times 10^{-12}\mathrm{C})}{1.2\mathrm{m}} = 1.499 \times 10^{-11} \mathrm{J}\) \(W_{3} = k \cdot (\frac{q_{1}q_{3}}{r_{1,3}} + \frac{q_{2}q_{3}}{r_{2,3}}) = (8.99 \times 10^{9} \mathrm{Nm^2/C^2}) \cdot ((\frac{(1.00 \times 10^{-12}\mathrm{C})(3.00 \times 10^{-12}\mathrm{C})}{1.2\mathrm{m}}) + (\frac{(2.00 \times 10^{-12}\mathrm{C})(3.00 \times 10^{-12}\mathrm{C})}{1.2\mathrm{m}})) = 4.498 \times 10^{-11} \mathrm{J}\) \(U_{total} = U_{1,2} + U_{1,3} + U_{2,3} = W_{2} + W_{3} = 1.499 \times 10^{-11} \mathrm{J} + 4.498 \times 10^{-11} \mathrm{J} = 5.997 \times 10^{-11} \mathrm{J}\) The work done to bring \(q_{1}\) to \(P\) is 0 J, \(q_{2}\) to \(Q\) is 1.499 x \(10^{-11}\) J, and \(q_{3}\) to \(R\) is 4.498 x \(10^{-11}\) J. The total potential energy stored in the final configuration of \(q_{1}\), \(q_{2}\), and \(q_{3}\) is 5.997 x \(10^{-11}\) J.

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Key Concepts

These are the key concepts you need to understand to accurately answer the question.

Work Done by Electric Forces
Understanding the concept of work done by electric forces is vital when analyzing electric charge interactions. Work is performed when a charge moves within an electric field, and it's analogous to lifting a weight against gravity. The work is calculated as the product of the force and the distance over which it acts, and it's dependent on the starting and ending points of the movement.

In electrostatics, the force exerted by one charge on another is an electric force. If a charge is moved closer to or further from another charge, work is done by or against this electric force. This is precisely what we consider when bringing a charge from infinity to a point within an electric field created by other charges. If no other charges are present, such as in the case of the first charge brought in from infinity, no work is done since no electric force acts on it.
Coulomb's Law
Central to the topic of electric forces is Coulomb's Law. This fundamental principle states that the magnitude of the electrostatic force between two point charges is directly proportional to the product of the magnitudes of the charges and inversely proportional to the square of the distance between them. The formula is expressed as:
\[ F = k \cdot \frac{|q_{1} \cdot q_{2}|}{r^2} \]
where \( F \) is the force, \( q_{1} \) and \( q_{2} \) are the charges, \( r \) is the distance between the charges, and \( k \) is Coulomb's constant. This law helps us calculate the amount of work done as charges are brought from infinity to a specific point in space when creating charge configurations.
Potential Energy in Electric Fields
The concept of potential energy in electric fields shares similarities with that of potential energy in gravitational fields. Just like a rock perched on a hill has gravitational potential energy due to its position, a charge in an electric field has electrical potential energy.

Electrostatic potential energy is the work done in moving a charge within an electric field without any acceleration. It's how we quantify the energy stored in the configuration of multiple charges. The potential energy depends on the relative positions of the charges involved, hence why the energy is zero when the first charge is brought in from infinity, as there's no pre-existing field to do work against.
Equilateral Triangle Charge Configuration
Our problem also considers a specific equilateral triangle charge configuration. This setup helps illustrate the symmetrical nature of forces and electric fields in a system of charges. With each of the three charges placed at the corners of an equilateral triangle, the distance between any two charges remains constant.

In such a configuration, the symmetry allows for easier calculations of work done to bring charges into position and the subsequent potential energy, using Coulomb's law. The total potential energy of the system is the sum of the potential energies of each unique pair of charges, an approach that simplifies the computation in scenarios with multiple charges in symmetrical patterns.

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Most popular questions from this chapter

A \(2.50-\mathrm{mg}\) dust particle with a charge of \(1.00 \mu \mathrm{C}\) falls at a point \(x=2.00 \mathrm{~m}\) in a region where the electric potential varies according to \(V(x)=\left(2.00 \mathrm{~V} / \mathrm{m}^{2}\right) x^{2}-\left(3.00 \mathrm{~V} / \mathrm{m}^{3}\right) x^{3} .\) With what acceleration will the particle start moving after it touches down?

Two metal balls of mass \(m_{1}=5.00 \mathrm{~g}\) (diameter \(=5.00 \mathrm{~mm}\) ) and \(m_{2}=8.00 \mathrm{~g}\) (diameter \(=8.00 \mathrm{~mm}\) ) have positive charges of \(q_{1}=5.00 \mathrm{nC}\) and \(q_{2}=8.00 \mathrm{nC}\), respectively. A force holds them in place so that their centers are separated by \(8.00 \mathrm{~mm}\). What will their velocities be after the force has been removed and they are separated by a large distance?

A proton with a speed of \(1.23 \cdot 10^{4} \mathrm{~m} / \mathrm{s}\) is moving from infinity directly toward a second proton. Assuming that the second proton is fixed in place, find the position where the moving proton stops momentarily before turning around.

A classroom Van de Graaff generator accumulates a charge of \(1.00 \cdot 10^{-6} \mathrm{C}\) on its spherical conductor, which has a radius of \(10.0 \mathrm{~cm}\) and stands on an insulating column. Neglecting the effects of the generator base or any other objects or fields, find the potential at the surface of the sphere. Assume that the potential is zero at infinity.

In which situation is the electric potential the highest? a) at a point \(1 \mathrm{~m}\) from a point charge of \(1 \mathrm{C}\) b) at a point \(1 \mathrm{~m}\) from the center of a uniformly charged spherical shell with a radius of \(0.5 \mathrm{~m}\) and a total charge of \(1 \mathrm{C}\) c) at a point \(1 \mathrm{~m}\) from the center of a uniformly charged rod with a length of \(1 \mathrm{~m}\) and a total charge of \(1 \mathrm{C}\) d) at a point \(2 \mathrm{~m}\) from a point charge of \(2 \mathrm{C}\) e) at a point \(0.5 \mathrm{~m}\) from a point charge of \(0.5 \mathrm{C}\)

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