Math Problem Statement
How many excess electrons must be added to an isolated spherical conductor 17.0 cm in diameter to produce an electric field of 5720 N/C just outside the surface?
Solution
To determine the number of excess electrons required, we follow these steps:
Step 1: Use Gauss’s Law
Gauss’s law states that the electric field just outside a conducting sphere is given by:
where:
- N/C (given electric field)
- is the total charge on the sphere
- is the radius of the sphere
- C²/N·m² (permittivity of free space)
Step 2: Solve for
The radius of the sphere is:
Rearrange Gauss's law to solve for :
Substituting values:
Let's calculate . math
Given values
epsilon_0 = 8.85e-12 # C²/N·m² R = 8.5e-2 # meters E = 5720 # N/C
Compute charge Q
Q = 4 * math.pi * epsilon_0 * R**2 * E QThe total charge required on the sphere is approximately C.
Step 3: Find the Number of Excess Electrons
Since the charge of a single electron is:
The number of excess electrons is:
Let's compute . Charge of a single electron e = 1.602e-19 # C
Number of excess electrons
n = Q / e nThe number of excess electrons required is approximately electrons.
Would you like a breakdown of any step?
Related Questions:
- How would the number of excess electrons change if the electric field were doubled?
- What would happen to the electric field if the radius of the sphere were increased while keeping the charge the same?
- How does Gauss’s law apply to a non-conducting sphere instead of a conductor?
- What would be the surface charge density on the sphere?
- How does the presence of a nearby conductor affect the charge distribution?
Tip:
In electrostatics, conductors in equilibrium have their excess charge residing entirely on the surface, making Gauss’s law very useful for spherical charge distributions!
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Math Problem Analysis
Mathematical Concepts
Electric Fields
Gauss's Law
Charge and Coulomb's Law
Basic Algebra
Formulas
E = Q / (4πε₀R²)
Q = 4πε₀R²E
n = Q / e
Theorems
Gauss's Law
Suitable Grade Level
Grades 10-12
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