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:

E=Q4πϵ0R2E = \frac{Q}{4\pi\epsilon_0 R^2}

where:

  • E=5720E = 5720 N/C (given electric field)
  • QQ is the total charge on the sphere
  • RR is the radius of the sphere
  • ϵ0=8.85×1012\epsilon_0 = 8.85 \times 10^{-12} C²/N·m² (permittivity of free space)

Step 2: Solve for QQ

The radius of the sphere is:

R=diameter2=17.0 cm2=8.5×102 mR = \frac{\text{diameter}}{2} = \frac{17.0 \text{ cm}}{2} = 8.5 \times 10^{-2} \text{ m}

Rearrange Gauss's law to solve for QQ:

Q=4πϵ0R2EQ = 4\pi\epsilon_0 R^2 E

Substituting values:

Q=4π(8.85×1012)(8.5×102)2(5720)Q = 4\pi (8.85 \times 10^{-12}) (8.5 \times 10^{-2})^2 (5720)

Let's calculate QQ. 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 4.60×1094.60 \times 10^{-9} C.

Step 3: Find the Number of Excess Electrons

Since the charge of a single electron is:

e=1.602×1019 Ce = 1.602 \times 10^{-19} \text{ C}

The number of excess electrons is:

n=Qe=4.60×1091.602×1019n = \frac{Q}{e} = \frac{4.60 \times 10^{-9}}{1.602 \times 10^{-19}}

Let's compute nn. 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 2.87×10102.87 \times 10^{10} electrons.

Would you like a breakdown of any step?

Related Questions:

  1. How would the number of excess electrons change if the electric field were doubled?
  2. What would happen to the electric field if the radius of the sphere were increased while keeping the charge the same?
  3. How does Gauss’s law apply to a non-conducting sphere instead of a conductor?
  4. What would be the surface charge density on the sphere?
  5. 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