Problem 101
Question
In a follow-up experiment, a charge of \(+40\) pC was placed at the center of an artificial flower at the end of a 30-cm long stem. Bees were observed to approach no closer than 15 cm from the center of this flower before they flew away. This observation suggests that the smallest external electric field to which bees may be sensitive is closest to which of these values? (a) \(2.4 \space N/C; (b) 16 \space N/C; (c) 2.7 \times 10^{-10} \space N/C; (d) 4.8 \times 10^{-10} \space N/C.\)
Step-by-Step Solution
Verified Answer
The smallest external electric field is 16 N/C.
1Step 1: Understand the Problem
We need to determine the smallest external electric field that deters the bee from approaching closer than 15 cm to a point charge of +40 pC. The options are given in N/C.
2Step 2: Identify Key Equations
We can calculate the electric field at a certain distance from a point charge using Coulomb's law: \[ E = \frac{k \cdot |q|}{r^2} \] where \( E \) is the electric field, \( k \) is Coulomb's constant \( (8.99 \times 10^9 \, \text{N m}^2/\text{C}^2) \), \( |q| \) is the magnitude of the charge, and \( r \) is the distance from the charge.
3Step 3: Insert Known Values
Here, \( |q| = 40 \, \text{pC} = 40 \times 10^{-12} \, \text{C} \) and \( r = 15 \, \text{cm} = 0.15 \, \text{m} \). Substitute these into the equation: \[ E = \frac{8.99 \times 10^9 \, \text{N m}^2/\text{C}^2 \cdot 40 \times 10^{-12} \, \text{C}}{(0.15 \, \text{m})^2} \]
4Step 4: Perform the Calculation
Calculate the electric field: \[ E = \frac{8.99 \times 10^9 \, \text{N m}^2/\text{C}^2 \cdot 40 \times 10^{-12} \, \text{C}}{0.0225 \, \text{m}^2} \] \[ E = \frac{359.6 \times 10^{-3} \, \text{N m}^2/\text{C}}{0.0225 \, \text{m}^2} \] \[ E \approx 15.98 \, \text{N/C} \]
5Step 5: Compare with Given Options
The calculated electric field \( 15.98 \, \text{N/C} \) is closest to option (b) 16 N/C.
Key Concepts
Coulomb's LawPoint ChargeElectric Field Calculation
Coulomb's Law
Coulomb's law is a fundamental principle in physics that describes the electrostatic interaction between two point charges. It states that the electric force between two charges is directly proportional to the product of the magnitudes of the charges and inversely proportional to the square of the distance between them. This law is represented mathematically as:
- Force (\[ F \]) between two charges: \[ F = \frac{k \cdot |q_1 q_2|}{r^2} \]
- \( k \) is Coulomb's constant (\( 8.99 \times 10^9 \, \text{N m}^2/\text{C}^2 \))
- \( |q_1| \) and \( |q_2| \) are the magnitudes of the charges
- \( r \) is the distance between the centers of the two charges
Point Charge
A point charge is an idealized model of a charged particle in which all the charge is concentrated at a single point in space. This simplification allows for straightforward calculations of electric fields and forces, as it reduces complex charge distributions to a simpler, ideal scenario. In physics problems like the one described, point charges serve as a useful approximation when the dimensions of the charged body are much smaller than the distances involved in the problem. This concept is crucial for applying Coulomb's law and determining the electric fields produced by such charges.When dealing with point charges, the electric field created by a point charge can be easily calculated at any given distance using the formula:
- Electric field (\[ E \]) due to a point charge: \[ E = \frac{k \cdot |q|}{r^2} \]
- \( |q| \) is the magnitude of the point charge
- \( r \) is the distance from the charge to the point where the field is being calculated
Electric Field Calculation
The electric field is a vector field that represents the force exerted by a charged object on a test charge placed within the field. To determine the strength of an electric field produced by a point charge, we use the equation derived from Coulomb's law:
- \[ E = \frac{k \cdot |q|}{r^2} \]
- Charge \( |q| = 40 \, \text{pC} = 40 \times 10^{-12} \, \text{C} \)
- Distance \( r = 15 \, \text{cm} = 0.15 \, \text{m} \)
Other exercises in this chapter
Problem 97
Two thin rods of length \(L\) lie along the \(x\)-axis, one between \(x = \frac{1}{2} a\) and \(x = \frac{1}{2} a + L\) and the other between \(x = -\frac{1}{2}
View solution Problem 99
What is the best explanation for the observation that the electric charge on the stem became positive as the charged bee approached (before it landed)? (a) Beca
View solution Problem 93
Two small spheres, each carrying a net positive charge, are separated by \(0.400 m\). You have been asked to perform measurements that will allow you to determi
View solution