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Chapter 20: Problem 26

The electric potential \(1.1 \mathrm{m}\) from a point charge \(q\) is \(2.8 \times 10^{4} \mathrm{V} .\) What is the value of \(q ?\)

Short Answer

Expert verified
The point charge \( q \) is approximately \( 3.43 \times 10^{-6} \text{ C} \).

Step by step solution

01

Understand the Formula for Electric Potential

The electric potential \( V \) due to a point charge \( q \) at a distance \( r \) is given by the formula: \[ V = \frac{kq}{r} \] where \( k \) is Coulomb's constant, approximately \( 8.99 \times 10^9 \text{ N m}^2/\text{C}^2 \). We need to solve this formula for \( q \).
02

Rearrange the Formula to Solve for q

Rearrange the formula \( V = \frac{kq}{r} \) to find \( q \). Multiply both sides by \( r \) and then divide by \( k \): \[ q = \frac{Vr}{k} \] This will allow us to calculate the charge \( q \).
03

Substitute Known Values

Substitute \( V = 2.8 \times 10^4 \text{ V} \), \( r = 1.1 \text{ m} \), and \( k = 8.99 \times 10^9 \text{ N m}^2/\text{C}^2 \) into the equation: \[ q = \frac{(2.8 \times 10^4)(1.1)}{8.99 \times 10^9} \]
04

Perform the Calculation

Calculate \( q \) by computing the numerator and then dividing by the denominator: \[ q = \frac{3.08 \times 10^4}{8.99 \times 10^9} \] Now, compute this value: \[ q \approx 3.43 \times 10^{-6} \text{ C} \]
05

State the Final Answer

The value of the point charge \( q \) is approximately \( 3.43 \times 10^{-6} \text{ C} \).

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Key Concepts

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

Point Charge
A point charge refers to a charge that is concentrated at a single point in space. It serves as an idealization to simplify our understanding of electric fields and potentials. A point charge
  • is often used to model small charged objects, assuming that their size is negligible compared to the distances involved in the problem.
  • enables straightforward calculations of electric potential and field by assuming the charge is concentrated at a point, which eliminates the need for complex integrations or calculus around its distribution.
When dealing with point charges, concepts like electric potential become much more manageable, allowing us to apply simple formulas and solve problems quickly. For instance, if you know the location of a point charge, you can determine its electric potential at any point around it.
Coulomb's Constant
Coulomb's constant, often represented as \( k \), is a fundamental value in electrostatics. It quantifies the force exerted between two point charges in a vacuum. This constant arises from Charles-Augustin de Coulomb's law, which describes how charged objects interact.
  • The typical value is approximately \( 8.99 \times 10^9 \text{ N m}^2/ ext{C}^2 \).
  • This constant is crucial for solving problems involving point charges, especially when calculating electric potential or the force between charges.
  • Coulomb's constant reflects the strength of electrical forces, which can be incredibly strong compared to gravitational forces, for example.
Having a consistent constant like \( k \) allows physicists and engineers to predictably solve problems involving charges in theoretical scenarios and has practical applications in technology and electronic devices.
Electric Potential Formula
The electric potential formula is essential for calculating the potential at a specific point caused by a point charge. The formula is \( V = \frac{kq}{r} \), where:
  • \( V \) is the electric potential.
  • \( k \) is Coulomb's constant.
  • \( q \) represents the charge.
  • \( r \) is the distance from the charge to the point where the potential is being calculated.
This formula shows that the potential depends directly on the magnitude of the point charge and inversely on the distance from it. For practical calculations like the one given in the exercise, simply rearranging this formula allows you to solve for the unspecified variable, such as the charge \( q \), when other values are known. This provides a clear method to understand the interaction between electric charges and fields.
Distance from Charge
The distance from a charge, denoted as \( r \) in the electric potential formula, plays a crucial role in determining the potential at a point.
  • As the distance \( r \) increases, the electric potential \( V \) decreases. This is because electric forces are strongest close to the source charge and diminish with distance.
  • Small changes in distance can significantly affect the potential due to its inverse relationship with \( r \) in the equation \( V = \frac{kq}{r} \).
  • By understanding the role of distance, you can predict how the electric potential varies in space around a point charge, which is especially important in fields such as electronics and physics.
In practical scenarios, assessing the electric potential at various distances helps in designing circuits, understanding electric fields, and even in predicting the behavior of charged particles in physics experiments.

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Most popular questions from this chapter

Cell Membranes The membrane of a living cell can be approximated by a parallel-plate capacitor with plates of area \(4.75 \times 10^{-9} \mathrm{m}^{2},\) a plate separation of \(8.5 \times 10^{-9} \mathrm{m},\) and a dielectric with a dielectric constant of \(4.5 .\) (a) What is the energy stored in such a cell membrane if the potential difference across it is \(0.0725 \mathrm{V}\) ? (b) Would your answer to part (a) increase, decrease, or stay the same if the thickness of the cell membrane is increased? Explain.

Find the electric energy density between the plates of a \(225 \cdot \mu F\) parallel-plate capacitor. The potential difference between the plates is \(345 \mathrm{V}\), and the plate separation is \(0.223 \mathrm{mm}\).

The Sodium Pump Living cells actively "pump" positive sodium ions \(\left(\mathrm{Na}^{7}\right)\) from inside the cell to outside the cell. This process is referred to as pumping because work must be done on the ions to move them from the negatively charged inner surface of the membrane to the positively charged outer surface. Given that the electric potential is \(0.070 \mathrm{V}\) higher outside the cell than inside the cell, and that the cell membrane is \(0.10 \mu \mathrm{m}\) thick (a) calculate the work that must be done (in joules) to move one sodium ion from inside the cell to outside. (b) If the thickness of the cell membrane is increased, does your answer to part (a) increase, decrease, or stay the same? Explain. (It is estimated that as much as \(20 \%\) of the energy we consume in a resting state is used in operating this "sodium pump.")

A parallel-plate capacitor has plates with an area of \(0.012 \mathrm{m}^{2}\) and a separation of \(0.88 \mathrm{mm} .\) The space between the plates is filled with a dielectric whose dielectric constant is \(2.0 .\) (a) What is the potential difference between the plates when the charge on the capacitor plates is \(4.7 \mu \mathrm{C}\) ? (b) Will your answer to part (a) increase, decrease, or stay the same if the dielectric constant is increased? Explain. (c) Calculate the potential difference for the case where the dielectric constant is \(4.0 .\)

Point charges \(+4.1 \mu C\) and \(-2.2 \mu C\) are placed on the \(x\) axis at \((11 \mathrm{m}, 0)\) and \((-11 \mathrm{m}, 0),\) respectively. (a) Sketch the electric potential on the \(x\) axis for this system. (b) Your sketch should show one point on the \(x\) axis between the two charges where the potential vanishes. Is this point closer to the \(+4.1-\mu \mathrm{C}\) charge or closer to the \(-2.2-\mu \mathrm{C}\) charge? Explain. (c) Find the point referred to in part (b).
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