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Chapter 17: Problem 36

At a distance of \(16 \mathrm{~m}\) from a charged particle, the electric field has a magnitude of \(100 \mathrm{~N} / \mathrm{C}\). (a) At what distance is the electric field \(400 \mathrm{~N} / \mathrm{C} ?\) (b) At what distance is it \(10 \mathrm{~N} / \mathrm{C} ?\)

Short Answer

Expert verified
8 meters for 400 N/C; approximately 50.56 meters for 10 N/C.

Step by step solution

01

Understanding the Problem

We need to find the distances at which the magnitude of the electric field due to a charged particle changes to specific values, namely 400 N/C and 10 N/C, while initially it is given as 100 N/C at 16 meters.
02

Understanding Electric Field Formula

The electric field due to a point charge is given by the formula \( E = \frac{k \cdot |q|}{r^2} \) where \( E \) is the electric field, \( k \) is Coulomb's constant, \( q \) is the charge, and \( r \) is the distance from the charge. For both parts, \( E \) changes but the same point charge is involved.
03

Using Inverse Square Law Relationship

When the electric field changes, since \( E \propto \frac{1}{r^2} \), it implies the relationship \( E_1 \cdot r_1^2 = E_2 \cdot r_2^2 \). Thus, \( r_2 = \sqrt{\frac{E_1}{E_2}} \cdot r_1 \), which we use to find the new distances.
04

Solving for 400 N/C

Substitute 100 N/C for \( E_1 \), 16 m for \( r_1 \), and 400 N/C for \( E_2 \) in the formula \( r_2 = \sqrt{\frac{E_1}{E_2}} \cdot r_1 \). This gives: \( r_2 = \sqrt{\frac{100}{400}} \cdot 16 = \sqrt{\frac{1}{4}} \cdot 16 = \frac{1}{2} \cdot 16 \). Therefore, \( r_2 = 8 \) meters.
05

Solving for 10 N/C

Substitute 100 N/C for \( E_1 \), 16 m for \( r_1 \), and 10 N/C for \( E_2 \) in the formula \( r_2 = \sqrt{\frac{E_1}{E_2}} \cdot r_1 \). This gives: \( r_2 = \sqrt{\frac{100}{10}} \cdot 16 = \sqrt{10} \cdot 16 \approx 3.16 \cdot 16 \approx 50.56 \) meters.
06

Conclusion

Thus, the distances where the electric field magnitudes are 400 N/C and 10 N/C are 8 meters and approximately 50.56 meters, respectively.

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Key Concepts

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

Inverse Square Law
The inverse square law is a fundamental concept in understanding how forces and fields behave over distance. It states that a physical quantity or intensity is inversely proportional to the square of the distance from the source. In the context of electric fields produced by point charges, this means:
  • The strength of the electric field decreases with the square of the distance from the charge. If you double the distance, the field strength becomes one-fourth as strong.
  • Mathematically, this is expressed as: \( E \propto \frac{1}{r^2} \), where \( E \) is the electric field and \( r \) is the distance from the charge.
Understanding this law helps predict how changes in distance affect the electric field's magnitude. This relationship is crucial when calculating fields at different distances from a point charge.
Point Charge
A point charge is an idealized model of a charged particle in physics. It simplifies calculations by assuming all the charge is concentrated at a single point, which allows physicists and students to apply basic principles of electrostatics without complex geometry.
  • Point charges provide a model for understanding electric fields and forces at a fundamental level.
  • This concept is critical when considering the effect of a charge on objects at various distances and how the electric field behaves around such a charge.
  • Even though real-world charges are distributed over a volume, using the point charge model often simplifies calculations significantly.
In the exercise, we assume our charged particle behaves like a point charge to calculate its electric field at different distances.
Coulomb's Constant
Coulomb鈥檚 constant, often represented by \( k \), is a fundamental constant in electrostatics, providing the proportionality factor in Coulomb鈥檚 Law. It defines the electric force between two point charges. The value of Coulomb's constant is approximately \( 8.99 \times 10^9 \text{ N m}^2/ ext{C}^2 \).
  • Coulomb鈥檚 constant facilitates the calculation of electric forces by ensuring that the units are consistent and physically meaningful.
  • In the electric field formula \( E = \frac{k \cdot |q|}{r^2} \), \( k \) helps relate the field's intensity to the charge \( q \) and distance \( r \).
  • Understanding \( k \) is key to solving problems related to electric fields and potential energies of charged objects.
Without Coulomb鈥檚 constant, calculations involving electric fields would lack the precision needed in physical models.
Electric Field Formula
The electric field formula is crucial for calculating the strength of an electric field generated by a point charge. It is expressed as \( E = \frac{k \cdot |q|}{r^2} \).
  • Here, \( E \) represents the electric field's magnitude at a distance \( r \) from the charge \( q \).
  • The formula incorporates Coulomb's constant \( k \), making it essential to accurately predict the field's behavior over distance.
  • In practical applications, this formula lets us determine how the field changes with distance, as it relies on the inverse square law.
In the given exercise, this formula was used to relate changes in the electric field's magnitude with corresponding distances by keeping the charge constant. This technique illustrates the practical utility of the formula in understanding electrostatic scenarios around point charges.

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Most popular questions from this chapter

An irregular neutral conductor has a hollow cavity inside of it and is insulated from its surroundings. An excess charge of \(+16 \mathrm{nC}\) is sprayed onto this conductor. (a) Find the charge on the inner and outer surfaces of the conductor. (b) Without touching the conductor, a charge of \(-11 \mathrm{nC}\) is inserted into the cavity through a small hole in the conductor. Find the charge on the inner and outer surfaces of the conductor in this case.

In a rectangular coordinate system, a positive point charge \(q=6.00 \mathrm{nC}\) is placed at the point \(x=+0.150 \mathrm{~m}, y=0,\) and an identical point charge is placed at \(x=-0.150 \mathrm{~m}, y=0 .\) Find the \(x\) and \(y\) components and the magnitude and direction of the electric field at the following points: (a) the origin; (b) \(x=0.300 \mathrm{~m}, y=0 ;\) (c) \(x=0.150 \mathrm{~m}, y=-0.400 \mathrm{~m} ;\) (d) \(x=0, y=0.200 \mathrm{~m}\)

Two very large parallel sheets of the same size carry equal magnitudes of charge spread uniformly over them, as shown in Figure \(17.58 .\) In each of the cases that follow, sketch the net pattern of electric field lines in the region between the sheets, but far from their edges. (Hint: First sketch the field lines due to each sheet, and then add these fields to get the net field.) (a) The top sheet is positive and the bottom sheet is negative, as shown, (b) both sheets are positive, (c) both sheets are negative.

You have a pure (24-karat) gold ring with mass \(17.7 \mathrm{~g}\). Gold has an atomic mass of \(197 \mathrm{~g} / \mathrm{mol}\) and an atomic number of \(79 .\) (a) How many protons are in the ring, and what is their total positive charge? (b) If the ring carries no net charge, how many electrons are in it?

A total charge of magnitude \(Q\) is distributed uniformly within a thick spherical shell of inner radius \(a\) and outer radius \(b\). (a) Use Gauss's law to find the electric field within the cavity \((r \leq a) .(\mathrm{b})\) Use Gauss's law to prove that the electric field outside the shell \((r \geq b)\) is exactly the same as if all the charge were concentrated as a point charge \(Q\) at the center of the sphere. (c) Explain why the result in part (a) for a thick shell is the same as that found in Example 17.10 for a thin shell. (Hint: A thick shell can be viewed as infinitely many thin shells.)
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