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\(\bullet\) \(\bullet\) The electric field due to a certain point charge has a magni- tude \(E\) at a distance of 1.0 \(\mathrm{cm}\) from the charge. (a) What will be the magnitude of this field (in terms of \(E\) ) if we move 1.0 \(\mathrm{cm}\) farther away from the charge? (b) What will be the magnitude of the field (in terms of \(E )\) if we move an additional 1.0 \(\mathrm{cm}\) far- ther away than in part (a)?

Short Answer

Expert verified
(a) \(\frac{E}{4}\); (b) \(\frac{E}{9}\).

Step by step solution

01

Understanding the Problem

We need to find the magnitude of the electric field due to a point charge at specified distances. The given distance where the magnitude is measured as \(E\) is \(1.0\, \text{cm}\). The task is to find the electric field magnitude at distances \(2.0\, \text{cm}\) and \(3.0\, \text{cm}\).
02

Electric Field Formula for a Point Charge

The electric field \(E\) due to a point charge is given by the formula:\[ E = \frac{kQ}{r^2} \]where \(k\) is Coulomb's constant, \(Q\) is the point charge, and \(r\) is the distance from the charge.
03

Expression for New Distance (Part a)

We are moving from \(1.0\, \text{cm}\) to \(2.0\, \text{cm}\). The new distance \(r' = 2.0\, \text{cm}\), so the electric field \(E'\) at this distance is:\[ E' = \frac{kQ}{(2r)^2} = \frac{kQ}{4r^2} \]
04

Relative Field Strength at 2.0 cm

We know that at \(r = 1.0\, \text{cm}\), the electric field is \(E = \frac{kQ}{r^2}\). Compare \(E'\) with \(E\):\[ E' = \frac{E}{4} \]Thus, at \(2.0\, \text{cm}\), the field is one-fourth of the field at \(1.0\, \text{cm}\).
05

Expression for New Distance (Part b)

We are now moving from \(2.0\, \text{cm}\) to \(3.0\, \text{cm}\). The new distance \(r'' = 3.0\, \text{cm}\), so the electric field \(E''\) at this distance is:\[ E'' = \frac{kQ}{(3r/2)^2} = \frac{kQ}{9r^2} \]
06

Relative Field Strength at 3.0 cm

We compare \(E''\) with \(E\):\[ E'' = \frac{E}{9} \]Thus, at \(3.0\, \text{cm}\), the field is one-ninth of the field at \(1.0\, \text{cm}\).

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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 is an idealized model of a charged object where the entire charge is assumed to be concentrated at a single point in space. In reality, charges are distributed over a volume, surface, or along a line, but for most practical purposes, especially in calculations involving electric fields, it simplifies things to consider the charge as existing at a point.

This concept is crucial in understanding electric fields because it allows us to calculate the electric field strength at various distances using consistent formulas. The simplification makes it easier to comprehend how charges interact at different points in space.
  • Point charges are theoretical; no physical size, only a position where charge is concentrated
  • Commonly used in problems to simplify the electric field calculations
  • Helps in visualizing forces and interactions in physics problems
Understanding the electric field around a point charge involves applying well-known principles, such as Coulomb's law, to determine how the field behaves as we move farther or closer to the charge.
Coulomb's Law
Coulomb's Law is fundamental to understanding electric fields and forces between charges. It describes how the force between two point charges is proportional to the product of their charges and inversely proportional to the square of the distance between them. This is mathematically expressed as: \[ F = rac{k |q_1 q_2|}{r^2} \] where:
  • \(F\) is the magnitude of the force between the charges,
  • \(q_1\) and \(q_2\) are the amounts of the charges,
  • \(r\) is the distance between the centers of the two charges, and
  • \(k\) is Coulomb's constant \( (8.99 \, \times \, 10^9 \, \text{N m}^2/\text{C}^2)\).
Coulomb's Law not only helps us calculate the force between two charges, it also informs us about the electric field created by a single charge: \[ E = rac{kQ}{r^2} \] Here, \(E\) is the electric field strength, \(Q\) is the charge producing the field, and \(r\) is the distance from the charge. This formula is crucial for solving electric field problems like the one given in our exercise.
Distance Dependence
The electric field from a point charge diminishes as we move away from the charge due to its distance dependence. In point charge scenarios, the electric field is directly influenced by the distance we are from the charge.

In mathematical terms, this distance dependence is represented by the variable \(r\) in the formula: \[ E = \frac{kQ}{r^2} \] As \(r\) increases, \(E\) decreases, showing how sensitive the field strength is to changes in distance. Specifically:
  • Double the distance \(r\), and the electric field \(E\) is reduced to one-fourth.
  • Triple the distance, and \(E\) reduces to one-ninth.
This shows that electric field strength weakens rapidly as one moves away from the charge. Understanding this helps explain the rationale behind the results in our specific exercise.
Inverse Square Law
The inverse square law is a critical principle that describes how a physical quantity (in this case, the electric field) decays with the square of the distance from its source. This principle not only applies to electric fields but also to gravitational fields, light, and sound intensity.

In our context, for electric fields, this law can be observed in the equation: \[ E = \frac{kQ}{r^2} \] As the distance \(r\) doubles, the field strength \(E\) becomes one-fourth because of the \(r^2\) term in the denominator. Thus, the field intensity decreases with the square of the increase in distance, illustrating that:
  • Tripling distance brings down field to one-ninth.
  • This quick reduction evidences why charges seem less influential with increasing distance.
The inverse square law helps explain why, at distances \(2.0 \, \text{cm}\) and \(3.0 \, \text{cm}\), the electric field is \( \frac{E}{4} \) and \( \frac{E}{9} \), respectively. It's a fundamental concept in physics needed to successfully analyze electric fields around point charges.

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Most popular questions from this chapter

\(\bullet$$\bullet\) In a certain region of space, the electric field \(E\) is uniform; i.e., neither its direction nor its magnitude changes in the region. (a) Use Gauss's law to prove that this region of space must be electrically neutral; that is, there must be no charge in this region. (b) Is the converse true? That is, in a region of space where there is no charge, must \(\vec{E}\) be uniform? Explain.

\(\bullet\) A proton is traveling horizontally to the right at \(4.50 \times 10^{6} \mathrm{m} / \mathrm{s}\) (a) Find the magnitude and direction of the weakest electric field that can bring the proton uniformly to rest over a distance of 3.20 \(\mathrm{cm} .\) (b) How much time does it take the proton to stop after entering the field? (c) What mini- mum field (magnitude and direction) would be needed to stop an electron under the conditions of part (a)?

\(\bullet$$\bullet \mathrm{A}-5.00 \mathrm{nC}\) point charge is on the \(x\) axis at \(x=1.20 \mathrm{m} . \mathrm{A}\) second point charge \(Q\) is on the \(x\) axis at \(-0.600 \mathrm{m} .\) What must be the sign and magnitude of \(Q\) for the resultant electric field at the origin to be (a) 45.0 \(\mathrm{N} / \mathrm{C}\) in the \(+x\) direction, \((\mathrm{b}) 45.0 \mathrm{N} / \mathrm{C}\) in the \(-x\) direction?

\(\bullet$$\bullet\) An electron is released from rest in a uniform electric field. The electron accelerates vertically upward, traveling 4.50 \(\mathrm{m}\) in the first 3.00\(\mu\) s after it is released. (a) What are the magnitude and direction of the electric field? (b) Are we justified in ignor- ing the effects of gravity? Justify your answer quantitatively.

\(\bullet\) A small object carrying a charge of \(-8.00 \mathrm{nC}\) is acted upon by a downward force of 20.0 \(\mathrm{nN}\) when placed at a certain point in an electric field. (a) What are the magnitude and direction of the electric field at the point in question? (b) What would be the magnitude and direction of the force acting on a proton placed at this same point in the electric field'?

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