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Two point charges, \(+3.40 \mu \mathrm{C}\) and \(-6.10 \mu \mathrm{C},\) are separated by \(1.20 \mathrm{~m} .\) What is the electric potential midway between them?

Short Answer

Expert verified
The total electric potential at the midpoint is \(-40,517 \text{ V}.\)

Step by step solution

01

Understand the Problem

We have two point charges, one positive and one negative, separated by a certain distance. We aim to find the electric potential at the midpoint of their separation. The electric potential due to a point charge is given by the formula \( V = \frac{kQ}{r} \), where \( k \) is the Coulomb's constant \( 8.99 \times 10^9 \text{ N m}^2/\text{C}^2 \), \( Q \) is the charge, and \( r \) is the distance from the charge.
02

Locate the Midpoint

The midpoint is exactly halfway between the two charges, which means it is \( \frac{1.20}{2} = 0.60 \text{ m} \) away from each charge.
03

Calculate Potential from +3.40 µC Charge

Use the formula \( V = \frac{kQ}{r} \) for the positive charge: \(V_+ = \frac{8.99 \times 10^9 \times 3.40 \times 10^{-6}}{0.60} = 50,965 \text{ V}.\)
04

Calculate Potential from -6.10 µC Charge

Use the formula \( V = \frac{kQ}{r} \) for the negative charge:\(V_- = \frac{8.99 \times 10^9 \times (-6.10 \times 10^{-6})}{0.60} = -91,482 \text{ V}.\)
05

Add the Potentials

The potential at the midpoint is the algebraic sum of the potentials due to each charge:\( V_{\text{total}} = V_+ + V_- = 50,965 - 91,482 = -40,517 \text{ V}. \)

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Key Concepts

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

Point Charges
Point charges are objects with a certain amount of electric charge concentrated at a single point in space. This simplification is often used in physics to make calculations easier. In many problems, like the one with our two charges of \(+3.40 \mu C\) and \(-6.10 \mu C\), these charges are considered to be points. This means:
- They don't occupy any space.
- The entire charge is concentrated at a geometrical point.
Point charges can be positive or negative, meaning they either have an excess of protons or electrons. The force and potential created by point charges are key to understanding many principles in electromagnetism, such as attraction and repulsion between charged objects.
Coulomb's Constant
Coulomb's constant, often denoted as \(k\), is a crucial part of calculating the forces and potential related to electric charges. It is a proportionality factor in Coulomb's law, which describes the electrostatic force between two point charges. Its value is approximately \(8.99 \times 10^9 \text{ N m}^2/\text{C}^2\).
  • This constant helps us understand how strong the electric force is between charges.
  • It appears in formulas that calculate electric potential and electric force.
Coulomb's law is expressed as \( F = \frac{k |Q_1 Q_2|}{r^2} \), where \(F\) is the force, \(Q_1\) and \(Q_2\) are the charges, and \(r\) is the distance between the charges. The law and constant describe how electric forces act over distances, helping to model phenomena in chemistry, physics, and engineering.
Electric Potential Formula
The electric potential at a point in space due to a point charge is calculated with the formula \( V = \frac{kQ}{r} \), where:
- \(V\) is the electric potential.
- \(k\) is Coulomb's constant.
- \(Q\) is the charge creating the potential.
- \(r\) is the distance from the charge to the point.
This formula tells us how much potential energy a unit charge would have if placed at a specific point in the electric field.
This concept is especially important when calculating potentials in locations between multiple charges, like the midpoint between two charges, as seen in the original exercise. By summing the potentials from multiple charges, we can determine the total electric potential at that point. This helps understand how energy might move within electric fields.
Midpoint Calculation
A midpoint is exactly halfway between two points. In problems involving charges, calculating the midpoint between them is often necessary to find values like electric potential. To find the midpoint, you simply divide the distance by two.
For example, if two charges are \(1.20 \text{ m}\) apart, the midpoint is at \( \frac{1.20}{2} = 0.60 \text{ m} \) from each charge.
This calculation ensures that you measure equal distances from each charge. Once the midpoint is found, you can use it to perform further calculations, like determining the electric potential at that specific location by adding the potentials due to each charge. Understanding how to locate midpoints is a basic skill in physics and helps in analyzing symmetrical setups involving forces and potentials.

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Most popular questions from this chapter

When you walk across a rug on a dry day, your body can become electrified, and its electric potential can change. When the potential becomes large enough, a spark of negative charges can jump between your hand and a metal surface. A spark occurs when the electric field strength created by the charges on your body reaches the dielectric strength of the air. The dielectric strength of the air is \(3.0 \times 10^{6} \mathrm{~N} / \mathrm{C}\) and is the electric field strength at which the air suffers electrical breakdown. Suppose a spark \(3.0 \mathrm{~mm}\) long jumps between your hand and a metal doorknob. Assuming that the electric field is uniform, find the potential difference \(\left(V_{\mathrm{knob}}-V_{\text {hand }}\right)\) between your hand and the doorknob.

Identical \(+1.8 \mu \mathrm{C}\) charges are fixed to adjacent corners of a square. What charge (magnitude and algebraic sign) should be fixed to one of the empty corners, so that the total electric potential at the remaining empty corner is \(0 \mathrm{~V} ?\)

A positive point charge is surrounded by an equipotential surface \(A\), which has a radius of \(r_{A}\). A positive test charge moves from surface \(A\) to another equipotential surface \(B\), which has a radius of \(r_{B}\). In the process, the electric force does negative work. (a) Does the electric force acting on the test charge have the same or opposite direction as the displacement of the test charge? (b) Is \(r_{B}\) greater than or less than \(r_{A}\) ? Explain your answers. Problem The positive point charge is \(q=+7.2 \times 10^{-8} \mathrm{C}\), and the test charge is \(q_{0}=+4.5 \times 10^{-11} \mathrm{C}\). The work done as the test charge moves from surface \(A\) to surface \(B\) is \(W_{A B}=-8.1 \times 10^{-9} \mathrm{~J}\). The radius of surface \(A\) is \(r_{A}=1.8 \mathrm{~m}\). Find \(r_{B}\). Check to see that your answer is consistent with your answers to the Concept Questions.

Charges \(q_{1}\) and \(q_{2}\) are fixed in place, \(q_{2}\) being located at a distance \(d\) to the right of \(\mathrm{q}_{1} . \mathrm{A}\) third charge \(q_{3}\) is then fixed to the line joining \(q_{1}\) and \(q_{2}\) at a distance \(d\) to the right of \(q_{2}\). The third charge is chosen so the potential energy of the group is zero; that is, the potential energy has the same value as that of the three charges when they are widely separated. Determine the value for \(q_{3}\), assuming that (a) \(q_{1}=q_{2}=q\) and (b) \(q_{1}=q\) and \(q_{2}=-q .\) Express your answers in terms of \(q\)

Point \(A\) is located \(0.25 \mathrm{~m}\) away from a charge of \(-2.1 \times 10^{-9} \mathrm{C}\). Point \(B\) is located \(0.50 \mathrm{~m}\) away from the charge. What is the electric potential difference \(V_{B}-V_{A}\) between these two points?

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