When two electrons move towards each other electrostatic potential energy

When we let the electron move freely in an electric field, the electric potential of electron must decrease but the electron moves opposite to the field so do we say that it's electric potential is increasing?

When the electron moves freely in an electric field, the electric potential of the electron increases, not decreases, and its electrical potential energy decreases. This is because high and low electric potential is defined by the convention that the direction of an electric field is the direction of the force that a positive charge would experience if placed in the field.

Based on this convention, a positive test charge placed in a field generated by another fixed positive charge will move in the direction of the field generated by the fixed charge, i.e., away from the fixed positive charge losing both electrical potential and electrical potential energy. On the other hand, if an external force moves the positive test charge against the direction of the field, i.e., towards the fixed positive charge, the electrical potential and potential energy of the positive test charge increases. So the location of the fixed positive charge is by definition high potential.

Now, if an electron is placed in the field, the field will move the electron in a direction opposite the field moving it toward the fixed positive charge. Since the location of the fixed positive charge is defined as high potential based on convention, the electron is considered to gain potential. On the other hand, the electron clearly loses electrical potential energy.

Hope this helps.

When two electrons move towards each other electrostatic potential energy

Text Solution

decreasesincreasesremains samebecomes zero

Answer : B

Solution : The electron has negative charge. When an electron is approaching towards another electron, then due to same negative charge repulsive force is produced between them. So, to bring them closer a work is done against this repulsive force. This work is stored in the form of electrostatic potential energy. Thus, electrostatic potential energy of system increases. <br> Alternative Electrostatic potential energy of system of two electrons <br> `U=(1)/(4piepsi_(0))((-e)(-e))/(r)=(1)/(4piepsi_(0))(e^(2))/(r)` <br> Thus, as r decreases, potential energy increases.

when an electron is brought near another electron, the work has to be done against the repulsive force since like charges repel each other. This work done is stored in the form of electrostatic potential energy and hence the electric potential energy of the system increases.

Hope it helps!

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The foundation for understanding bonding in chemistry is a fairly simple concept: Potential energy. Potential energy (Ep) is the energy which one or more bodies have because of their position.

==Gravitational Potential Energy==IWC replica Réplica de reloj It's simple because we can understand it in terms of the common experience of lifting an object. Audemars Piguet Royal According to the law of conservation of energy, the muscle energy expended for a bird to rise from the surface of the earth to 100 m cannot be created or destroyed, so it must be converted to potential energy, Ep. The bird's Ep is large at 100 m height. In other words, the bird has a large amount of potential energy that can then be converted to other forms. As the bird descends, Ep is released, or better, converted to other forms of energy, like wind turbulance or heat in the atmosphere. The Potential energy is calculated with the formula:

\[\text{E}_p = \text{mgh}\label{1}\]

where g is the acceleration of gravity (9.8 m s-2, m is the mass, and h is the height of the object.

According to equation \(\ref{1}\), the PE must be zero when the height is zero. For a mass of 1 g (1/5 the mass of a nickel) at a height of 100 m:

\[\text{E}_p = \text{mgh} = \text{10}^{-3} \text{kg} \times \text{9.8 m s}^{-2} \times \text{10}^2 \text{m} = \text{0.98 J}\nonumber\]

Figure \(\PageIndex{1}\) Gravitational Attraction

When the object falls, it releases energy, and the energy is given a negative sign to show that it's lost by the object. It's change in energy, "ΔE", is always calculated as the final energy (Ep,f) minus the initial energy (Ep,i).

\[\Delta \text{E}_p = \text{E}_{p,f} - \text{E}_{p,i} = \text{0 J - 0.98 J} = -\text{0.98 J}\nonumber\]

Electrostatic forces between electrons and nuclei lead to chemical bonding, and the picture is a little different because these may be either forces of attraction or repulsion. Because of the electrostatic force of attraction or repulsion between them, two charged particles will vary in energy as we alter the distance between them.

Opposite (+ and -) charges attract, just like the earth and any object, so the potential energy increases as we put effort in to separate the charges. Once they're separated, they will "fall" together, releasing Ep.

But like charges (+ and +, or - and -) repel, so it takes effort to push them together, and when they're close together they have high potential energy. If they're pushed together and released, they will fly apart, releasing energy.

Because the potential energy of like-charged particles increases as they are brought closer together, while that of opposite-charged particles decreases, it is convenient to assign a value of zero potential energy to two charged particles which are a long distance apart. Bringing one positive and one negative particle together decreases Ep, giving a negative value, indicating that energy is released.

Figure \(\PageIndex{2}\) Electrostatic Attraction and Repulsion

Suppose we have charges of +1 and –1 μC separated by 1 cm, for example. The charges could be separated by hand, and by the time they're 100 m (about a football field) apart, their attractive force would be negligible (it's taken to be zero at infinite separation). Expenditure of muscle energy will be necessary to carry out such a separation. That is, because the charges attract each other, we must do work to pull them apart equivalent to 0.898 J[1].

\[\Delta \text{E}_p = \text{E}_{p,f} - \text{E}_{p,i} = \text{0 J} - \text{E}_{p,i} = +\text{0.898 J}\n\nonumber\]

Thus

\[-\text{E}_{p,i} = + \text{0.898 J - 0 J}\nonumber\]

\[\text{E}_{p,i} = -\text{0.898 J}\nonumber\]

We say that the 0.898 J is gained by the two charges and stored as potential energy. We can always regain this energy by reversing the process during which it was stored. If the two opposite charges are returned to their original separation of 1 cm, their potential energy will decrease by 0.898 J.

\[\Delta \text{E}_p = \text{E}_{p,f} - \text{E}_{p,i} = \text{-0.898 J-0 J} = -\text{0.898 J}\nonumber\]

The energy released will appear as kinetic energy, as heat, or in some other form, but it cannot be destroyed.

If we had taken two particles both of which had a charge of +1 μC for our example of potential energy, work would have been required to push them together against their repulsive force. Their potential energy would increase as they were brought together from the ends of a football field, and 0.898 J would be required to move them to a distance of 1 cm apart. Note that in the figure the higher potential energies are given positive values.

\[\Delta \text{E}_p = \text{E}_{p,f} - \text{E}_{p,i} = +\text{0.898 J - 0 J} = +\text{0.898 J}\nonumber\]

Bringing a pair of positive charges (or a pair of negative charges) closer together increases Ep to a positive value. Bringing one positive and one negative particle together decreases Ep, giving a negative value.

From ChemPRIME: 5.10: Potential Energy

References

  1. ↑ the formula is
    When two electrons move towards each other electrostatic potential energy

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