Problem 47 - College Physics

Question

\(\bullet\) \(\bullet\) 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)?

Step-by-Step Solution

Answer

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

Step 1: 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 \(3.0\, \text{cm}\).

Step 2: 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.

Step 3: 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:\[ E' = \frac{kQ}{(2r)^2} = \frac{kQ}{4r^2} \]

Step 4: 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 \(1.0\, \text{cm}\).

Step 5: 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:\[ E'' = \frac{kQ}{(3r/2)^2} = \frac{kQ}{9r^2} \]

Step 6: 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 \(1.0\, \text{cm}\).

Key Concepts

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.