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How much work would be done by an electric field in moving a proton from a point at a potential of \(+180 . \mathrm{V}\) to a point at a potential of \(-60.0 \mathrm{~V} ?\)

Short Answer

Expert verified
Answer: The work done by the electric field is -3.84 × 10^{-17} J.

Step by step solution

01

Identify the given information

We are given: - Potential at point A (initial point) is \(+180 \mathrm{V}\) - Potential at point B (final point) is \(-60.0 \mathrm{V}\) - Charge of a proton is \(+1.60 × 10^{-19} \mathrm{C}\)
02

Calculate the potential difference

To find the potential difference, we subtract the initial potential from the final potential: Potential difference = potential at point B - potential at point A Potential difference = \((-60.0 \mathrm{V}) - (+180 \mathrm{V})\) Potential difference = \(-240 \mathrm{V}\)
03

Calculate the work done by the electric field

Now we can use the formula for work done by an electric field: Work = charge × potential difference Work = \((1.60 × 10^{-19} \mathrm{C}) × (-240 \mathrm{V})\) Work = \(-3.84 × 10^{-17} \mathrm{J}\) The electric field does \(-3.84 × 10^{-17} \mathrm{J}\) of work in moving a proton from a point at a potential of \(+180 \mathrm{V}\) to a point at a potential of \(-60.0 \mathrm{V}\). The negative sign indicates that the work is done against the electric field.

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Key Concepts

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

Electric Field
Imagine being in a pool with waves moving around you, pushing you in different directions. An electric field is somewhat similar but in an electric context. It's an invisible influence around charged objects that exerts a force on other charges within its reach. Just as waves in a pool exert forces on you, the electric field exerts a force on other electric charges.

When a single positive charge is placed in this field, it will experience a push or pull. This force is due to the electric field, which can be thought of as an electric 'pressure' that permeates the space around charged objects. In more technical terms, it's usually represented by the symbol E and can be calculated using the formula \( E = F / q \), where F is the force experienced by a charge, and q is the magnitude of the charge.
Potential Difference
To understand the potential difference, consider a hill. Hiking to the top requires effort, and going back down will let gravity do the work for you. Now, imagine this in terms of electric charges. Potential difference (also called voltage) is the measure of how much 'electrical pressure' there is to move a charge from one point to another.

It’s the difference in energy that a charge will feel at two different points in an electric field. It is the 'height' of the hill that a charge must climb through the electric field. When calculating, as in the exercise, it is found by subtracting one potential from another. This concept is important because it helps to determine how much work will be done by or on a charge as it moves between two points within an electric field.
Work Done by Electric Field
When a force moves an object through a distance, work is done. In the field of electricity, the work done by an electric field refers to the energy needed to move a charge within an electric field. It's like the energy you expend when lifting a box off the ground or when an electric field 'lifts' a charge against its natural direction.

The work done is calculated by multiplying the charge (in coulombs) by the potential difference (in volts). The result is energy, measured in joules. In the example problem, once we know the potential difference and the electric charge of a proton, we can find out the work done by the electric field in moving the proton from a high potential point to a lower one.
Electric Charge
The concept of an electric charge is at the heart of all phenomena related to electricity. Think of charges as tiny packets of electric 'stuff' that can exert forces and create electric fields. There are two types of electrical charges, positive and negative, and they are the basic building blocks of electricity.

Like tiny magnets with 'north' and 'south' poles, opposite charges attract each other and like charges repel. The unit used to measure charge is the coulomb (C), which is a large unit so we often talk about charges in terms of microcoulombs (\(10^{-6}\) C) or even smaller, as is the case with the charge of a proton which is \(1.60 × 10^{-19}\) C. The movement of charges is the basis for electric currents, and it's the balancing and imbalances of these charges that lead to most electric phenomena.

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Most popular questions from this chapter

Two protons at rest and separated by \(1.00 \mathrm{~mm}\) are released simultaneously. What is the speed of either at the instant when the two are \(10.0 \mathrm{~mm}\) apart?

A proton gun fires a proton from midway between two plates, A and \(\mathrm{B}\), which are separated by a distance of \(10.0 \mathrm{~cm}\); the proton initially moves at a speed of \(150.0 \mathrm{~km} / \mathrm{s}\) toward plate \(\mathrm{B}\). Plate \(\mathrm{A}\) is kept at zero potential, and plate \(\mathrm{B}\) at a potential of \(400.0 \mathrm{~V}\). a) Will the proton reach plate \(\mathrm{B}\) ? b) If not, where will it turn around? c) With what speed will it hit plate \(\mathrm{A} ?\)

What would be the consequence of setting the potential at \(+100 \mathrm{~V}\) at infinity, rather than taking it to be zero there? a) Nothing; the field and the potential would have the same values at every finite point. b) The electric potential would become infinite at every finite point, and the electric field could not be defined. c) The electric potential everywhere would be \(100 \mathrm{~V}\) higher, and the electric field would be the same. d) It would depend on the situation. For example, the potential due to a positive point charge would drop off more slowly with distance, so the magnitude of the electric field would be less.

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?

The amount of work done to move a positive point charge \(q\) on an equipotential surface of \(1000 \mathrm{~V}\) relative to that done to move the charge on an equipotential surface of \(10 \mathrm{~V}\) is a) the same. b) less. c) more. d) dependent on the distance the charge moves.

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