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Chapter 19: Problem 13

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

Short Answer

Expert verified
The electric potential midway is approximately \\(-4.054 \\\\times 10^4 \\, \\\\text{V}\\).

Step by step solution

01

Understand the Concept of Electric Potential

Electric potential at a point is the work done in bringing a unit positive charge from infinity to that point without acceleration. The potential due to a point charge is given by the formula: \[ V = \frac{k \cdot q}{r} \]where \(k\) is Coulomb's constant \(8.99 \times 10^9 \, \text{N} \, \text{m}^2/\text{C}^2\), \(q\) is the charge, and \(r\) is the distance from the charge.
02

Identify Given Values

Identify the charges and the distances involved:\[ q_1 = +3.40 \, \mu\text{C} = 3.40 \times 10^{-6} \, \text{C} \]\[ q_2 = -6.10 \, \mu\text{C} = -6.10 \times 10^{-6} \, \text{C} \]The total distance between the charges is \(1.20 \, \text{m}\). Thus, the point midway between them is at \(0.60 \, \text{m}\) from each charge.
03

Calculate the Electric Potential Due to Each Charge

Calculate the potential at the midpoint due to each charge:1. Potential due to \(q_1\):\[ V_1 = \frac{8.99 \times 10^9 \cdot 3.40 \times 10^{-6}}{0.60} \, \text{V} = \frac{30.566 \times 10^3}{0.60} \, \text{V} \approx 5.094 \times 10^4 \, \text{V} \]2. Potential due to \(q_2\):\[ V_2 = \frac{8.99 \times 10^9 \cdot (-6.10) \times 10^{-6}}{0.60} \, \text{V} = -\frac{54.899 \times 10^3}{0.60} \, \text{V} \approx -9.148 \times 10^4 \, \text{V} \]
04

Add Potentials to Find the Net Electric Potential

The net electric potential at the midpoint is the algebraic sum of the potentials due to each charge:\[ V_{\text{net}} = V_1 + V_2 \]\[ V_{\text{net}} = 5.094 \times 10^4 \, \text{V} + (-9.148 \times 10^4 \, \text{V}) \]\[ V_{\text{net}} = -4.054 \times 10^4 \, \text{V} \]

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Key Concepts

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

Coulomb's Law
Coulomb's Law is the fundamental equation that quantifies the electric force between two charged objects. It helps us understand how charges interact at a distance. The law is expressed mathematically as \[ F = \frac{k \cdot |q_1 \cdot q_2|}{r^2} \]where:
  • \(F\) is the magnitude of the force between the charges.
  • \(k\) is Coulomb's constant, approximately \(8.99 \times 10^9 \, \text{N} \, \text{m}^2/\text{C}^2\).
  • \(q_1\) and \(q_2\) are the magnitudes of the charges.
  • \(r\) is the distance between the centers of the two charges.
The force is attractive if the charges are oppositely charged and repulsive if they have like charges. This principle sets the groundwork for calculating other properties such as electric potential and field.
Point Charge
A point charge is an idealized model of a charged object where the charge is considered to be concentrated at a single point in space. This simplification is helpful in theoretical calculations in physics.When dealing with a point charge,
key properties to consider include:
  • The charge magnitude (how much charge is there).
  • The algebraic sign (positive or negative), influencing the direction of interaction with other charges.
An important formula to know is the calculation of electric potential \(V\) due to a single point charge:\[ V = \frac{k \cdot q}{r} \]where \(V\) is the electric potential, \(q\) is the point charge, and \(r\) is the distance from the charge to the point where potential is being calculated.
This concept is crucial when solving problems related to electric potential, particularly where you must find contributions from multiple point charges.
Electric Field
The electric field is a vector field that surrounds electric charges. It represents the force that would be experienced by a positive test charge placed within the field. It is a useful concept for visualizing the influence of charges in space.
The strength and direction of the electric field are given by the formula:\[ E = \frac{F}{q} = \frac{k \cdot |Q|}{r^2} \]where:
  • \(E\) is the electric field magnitude.
  • \(F\) is the force experienced by the test charge \(q\).
  • \(Q\) is the source charge creating the field.
  • \(r\) is the distance from the charge \(Q\) producing the field.
The electric field points away from positive charges and towards negative charges. By understanding electric fields, we can better interpret how charges influence one another at a distance and how they contribute to forms of electric potential.

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Most popular questions from this chapter

An empty parallel plate capacitor is connected between the terminals of a \(9.0-\mathrm{V}\) battery and charged up. The capacitor is then disconnected from the battery, and the spacing between the capacitor plates is doubled. As a result of this change, what is the new voltage between the plates of the capacitor?

A positive charge \(q_{1}\) is located \(3.00 \mathrm{m}\) to the left of a negative charge \(q_{2} .\) The charges have different magnitudes. On the line through the charges, the net electric field is zero at a spot \(1.00 \mathrm{m}\) to the right of the negative charge. On this line there are also two spots where the total electric potential is zero. Locate these two spots relative to the negative charge.

Location \(A\) is \(3.00 \mathrm{m}\) to the right of a point charge \(q .\) Location \(B\) lies on the same line and is \(4.00 \mathrm{m}\) to the right of the charge. The potential difference between the two locations is \(V_{B}-V_{A}=45.0 \mathrm{V}\). What are the magnitude and sign of the charge?

The electronic flash attachment for a camera contains a capacitor for storing the energy used to produce the flash. In one such unit, the potential difference between the plates of an \(850-\mu \mathrm{F}\) capacitor is \(280 \mathrm{V}\). (a) Determine the energy that is used to produce the flash in this unit. (b) Assuming that the flash lasts for \(3.9 \times 10^{-3}\) s, find the effective power or "wattage" of the flash.

The electric potential energy stored in the capacitor of a defibrillator is 73 J, and the capacitance is \(120 \mu \mathrm{F}\). What is the potential difference that exists across the capacitor plates?
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