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Chapter 3: Problem 3

If \(d\) is finite, and the particles have the same charge (that is, \(q_{1}=q_{2}\) ), is \(V>0\) or is \(V<0 ?\) Explain your answer.

Short Answer

Expert verified
The potential difference \(V\) would be greater than zero (\(V > 0\)) since both charges are equal resulting in a force of repulsion. Since work is done against this repulsion to bring the two charges closer, the potential difference becomes positive.

Step by step solution

01

Understand the basics

Having equal charges \(q_{1} = q_{2}\) means that particles repel each other. The work done by an electrostatic field \(E\) to relocate a charge from a point \(A\) to another point \(B\) is the source of the potential difference \(V\) (also known as voltage). When work is done against the direction of the field, the potential difference is positive. Conversely, if work follows the direction of the field, the potential difference becomes negative.
02

Determine the potential difference

Since the charges are equal and positive, they repel each other. Therefore, to move one charge closer to the other, work is required against the electrostatic field, which leads to a positive potential difference (\(V > 0\)).

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Key Concepts

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

Electric Charge
Electric charge is a fundamental property of matter that quantifies an object's ability to exert electromagnetic force. It's the 'electricity' part in various phenomena such as static electricity or current flow. Charges come in two types: positive and negative. Like charges repel each other, while opposite charges attract.
For example, when you rub a balloon on your hair, it acquires an electric charge which then can stick to a wall. That's electric charge in action! Objects with the same charge, as in the given exercise (\(q_1 = q_2\)), repel each other because they are of like sign, confirming why charges play a key role in understanding electrostatic repulsion or attraction.
Electrostatic Field
An electrostatic field is a field around a charged object, where a force would be experienced by another charge placed within this field. It's a vector field which means it has both magnitude and direction; the direction is away from positive charges and towards negative charges.
Imagine you have a balloon with a positive charge, and you bring a positively charged ping-pong ball near it. The ping-pong ball will move away from the balloon because it is inside the balloon's electrostatic field, experiencing a repulsive force.
Voltage in Physics
Voltage, also known as electric potential difference, is the measure of electric potential energy per unit charge. It's the 'push' that gets charges moving in a circuit. You can think of it like water pressure in a pipe; it's what drives the water through. When referring to voltage in a static electric field, it’s the work done per unit charge to move a charge within the field.
In the case of our exercise problem, a positive voltage (\(V > 0\)) indicates that work needs to be done to bring the same charges closer together because they naturally repel each other, just like it would require effort to push water up against the flow.
Work and Energy in Electrostatics
In the context of electrostatics, work is done when a force moves a charge in the presence of an electrostatic field. The energy required to do this work is stored as electric potential energy.
Think of lifting a book off the floor onto a shelf; you're doing work against gravity, and the book has higher potential energy on the shelf. Similarly, moving charges against the electrostatic field increases their electric potential energy. In our textbook problem, the necessity of doing work to bring similarly charged particles together results in a positive increase in potential energy, leading to a positive voltage.

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Most popular questions from this chapter

Which of the following systems will have the larger ionization energy? Explain your reasoning. i) an electron at a distance \(d_{1}\) from a nucleus with charge \(+2\) ii) an electron at a distance \(d_{1}\) from a nucleus with charge \(+1\)

Consider a hydrogen atom and a helium ion, \(\mathrm{He}^{+}\). Which of these do you expect to have the larger ionization energy? Explain your reasoning, including any assumptions you make.

If \(q\) for an electron is \(-1\), a) what is \(q\) for a proton? b) what is \(q\) for a neutron? c) what is \(q\) for the nucleus of a \(C\) atom?

Assuming that \(q_{1}\) and \(q_{2}\) remain constant, what happens to the magnitude of \(\bar{V}\) if the separation, \(d\), is increased?

Recall that a \({ }^{1} \mathrm{H}\) atom consists of a proton as the nucleus and an electron outside of the nucleus. Is the potential energy, \(V\), of a hydrogen atom a positive or negative number? Explain your answer.
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