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Chapter 17: Problem 19

(I) A point charge \(Q\) creates an electric potential of +165 V at a distance of 15 cm. What is \(Q\)?

Short Answer

Expert verified
The charge \(Q\) is approximately \(2.75 \times 10^{-9} \, \text{C}\).

Step by step solution

01

Understand the Formula for Electric Potential

The electric potential (V) at a distance \( r \) from a point charge \( Q \) is given by the formula:\[V = \frac{kQ}{r}\]where \( k \) is Coulomb's constant \( 8.99 \times 10^9 \, \text{N m}^2/\text{C}^2 \).
02

Set Up the Known Values

Substitute the known values into the equation. We have:\( V = 165 \, \text{V} \),\( r = 0.15 \, \text{m} \) (convert 15 cm to meters),\( k = 8.99 \times 10^9 \, \text{N m}^2/\text{C}^2 \).
03

Rearrange the Formula to Solve for Q

Rearrange the formula \( V = \frac{kQ}{r} \) to solve for \( Q \):\[Q = \frac{V \, r}{k}\]
04

Plug the Values Into the Equation

Insert the values into the rearranged formula:\[Q = \frac{165 \, \text{V} \times 0.15 \, \text{m}}{8.99 \times 10^9 \, \text{N m}^2/\text{C}^2}\]
05

Calculate the Charge Q

Calculate \( Q \) using the numbers provided:\[Q = \frac{24.75}{8.99 \times 10^9}\]\( Q \approx 2.75 \times 10^{-9} \, \text{C} \).

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Key Concepts

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

Coulomb's constant
Coulomb's constant, often represented as the symbol \( k \), plays a crucial role in electric force and potential calculations. Its value is approximately \( 8.99 \times 10^9 \, \text{N} \cdot \text{m}^2/\text{C}^2 \). This constant derives from Coulomb's law, which describes the magnitude of the force between two charges.

Understanding Coulomb's law helps us grasp how \( k \) is used:
  • It establishes the relationship between electric force, charge magnitude, and distance.
  • \( k \) is a proportionality constant that helps calculate forces in electrostatic interactions.
  • In the context of electric potential, Coulomb's constant allows us to use the formula \( V = \frac{kQ}{r} \), where \( V \) is the electric potential created by a point charge.
Whether calculating electric forces or potential, always remember that \( k \) allows us to relate charge interaction to measurable quantities like force and potential.
Point charge
A point charge refers to an idealized model of a charged particle. It is assumed to have all its charge concentrated at a single point in space. This simplification is useful for theoretical calculations in physics.

In electric potential and field calculations:
  • A point charge provides a simple way to envision how an electric field originates from a charged object.
  • Despite being an abstraction, it accurately describes the behavior of small charged objects, like electrons, at certain distances.
  • The effects of a point charge diminish with distance, meaning the further away you are, the weaker the electric field it produces.
The concept of a point charge is foundational in electrostatics, helping us understand interactions and predict behaviors of real-world charged objects.
Electric charge calculation
Electric charge calculation involves determining the quantity of charge based on its effects, such as electric potential. With known values of potential, distance, and Coulomb's constant, we can find an unknown charge using the formula:

\[Q = \frac{V \times r}{k}\]There are a few steps to follow when solving problems involving electric charge calculations:
  • Identify and appropriately convert all given values, like distance in meters.
  • Rearrange the electric potential formula to solve for \( Q \) if necessary.
  • Substitute the known values into the equation to find the charge.
In our specific exercise, this method helped determine a point charge \( Q = 2.75 \times 10^{-9} \text{ C} \) that creates an electric potential of 165 volts at 0.15 meters away. Understanding the principles and steps makes it easier to solve similar problems efficiently.

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Most popular questions from this chapter

A huge 4.0-F capacitor has enough stored energy to heat 2.8 kg of water from 21\(^\circ\)C to 95\(^\circ\)C. What is the potential difference across the plates?

A parallel-plate capacitor with plate area \(A =\) 2.0 m\(^2\) and plate separation \(d =\) 3.0 mm is connected to a 35-V battery (Fig. 17-51a). (\(a\)) Determine the charge on the capacitor, the electric field, the capacitance, and the energy stored in the capacitor. (\(b\)) With the capacitor still connected to the battery, a slab of plastic with dielectric strength \(K =\) 3.2 is placed between the plates of the capacitor, so that the gap is completely filled with the dielectric (Fig. 17-51b).What are the new values of charge, electric field, capacitance, and the energy stored in the capacitor?

Near the surface of the Earth there is an electric field of about 150 V/m which points downward. Two identical balls with mass \(m =\) 0.670 kg are dropped from a height of 2.00 m, but one of the balls is positively charged with \(q_1 =\) 650\(\mu\)C and the second is negatively charged with \(q_2 =\) 650\(\mu\)C Use conservation of energy to determine the difference in the speed of the two balls when they hit the ground. (Neglect air resistance.)

(II) An uncharged capacitor is connected to a 21.0-V battery until it is fully charged, after which it is disconnected from the battery. A slab of paraffin is then inserted between the plates. What will now be the voltage between the plates?

(II) How strong is the electric field between the plates of a 0.80-\(\mu\)F air- gap capacitor if they are 2.0 mm apart and each has a charge of 62\(\mu\)C
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