Problem 50 - University Physics with Modern Physics

Question

22.50. (a) How many excess electrons must be distributed uni- formly within the volume of an isolated plastic sphere 30.0 \(\mathrm{cm}\) in diameter to produce an electric field of 1150 \(\mathrm{N} / \mathrm{C}\) just outside the surface of the sphere? (b) What is the electric field at a point 10.0 \(\mathrm{cm}\) outside the surface of the sphere?

Step-by-Step Solution

Answer

(a) \( n \approx 6.73 \times 10^{11} \) excess electrons. (b) \( E \approx 414 \) N/C.

Step 1: Understanding the Problem

We need to calculate the number of excess electrons needed to create a specified electric field just outside a charged sphere. The diameter of the sphere is 30 cm, which means the radius is 15 cm. The electric field just outside the surface is given as 1150 N/C.

Step 2: Calculating Charge Required

The electric field just outside a uniformly charged sphere is given by the formula \( E = \frac{kQ}{r^2} \), where \( k \) is Coulomb's constant \( 8.99 \times 10^9 \) N·m²/C², \( Q \) is the charge, and \( r \) is the radius of the sphere. Rearrange to solve for \( Q \): \( Q = \frac{Er^2}{k} \). Substitute \( E = 1150 \) N/C and \( r = 0.15 \) m to find \( Q \).

Step 3: Substituting Values

Substitute the values into the formula for \( Q \): \[ Q = \frac{1150 \times (0.15)^2}{8.99 \times 10^9} \]. Calculate \( Q \) to find the charge in coulombs.

Step 4: Converting Charge to Electrons

An electron has a charge of \( -1.6 \times 10^{-19} \) C. To find the number of electrons (\( n \)) that corresponds to the charge \( Q \), use \( n = \frac{Q}{1.6 \times 10^{-19}} \). Substitute \( Q \) from the previous step to find the value of \( n \).

Step 5: Calculating Electric Field at a Distance

For part (b), use \( E = \frac{kQ}{R^2} \) where \( R \) is the total distance from the sphere's center to the point of interest, which includes the sphere's radius plus 10 cm. Thus, \( R = 0.25 \) m. Substitute \( Q \) from part (a) and \( R \) into the formula to calculate \( E \).

Key Concepts

Electric Field

The electric field is a fundamental concept in electrostatics. It's a region around a charged object where other charges experience a force. For a charged sphere, the electric field just outside its surface behaves as if all the charge were concentrated at the center. This allows us to simplify calculations.
An electric field strength is expressed in newtons per coulomb (N/C). In this exercise, the electric field strength just outside the sphere is given as 1150 N/C, which helps us determine how many excess electrons need to be distributed on the sphere.

Coulomb's Law

Coulomb's Law provides a quantitative description of the electric force between charged objects. The law states that the 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.
The formula used is:

• \( F = \frac{k \cdot |Q_1 \cdot Q_2|}{r^2} \)

Where:

• \( k \) is Coulomb's constant, valued at \( 8.99 \times 10^9 \) N·m²/C².
• \( Q_1 \) and \( Q_2 \) are the charges.
• \( r \) is the distance between the charges.

While this exercise doesn't compute forces, it uses the principles of Coulomb's Law to understand electric fields' dependencies on charge and distance.

Charge Distribution

Charge distribution tells us how charge is spread over or within an object. In this scenario, an isolated plastic sphere is assumed to have a uniform charge distribution, making the electric field calculation straightforward. The charge is spread evenly over the sphere's surface, allowing the entire charge to be treated as if it were concentrated at the center for the purpose of calculating the electric field outside the sphere.
Uniform distribution simplifies many calculations and helps us use formulas involving radial distribution, crucial for this step-by-step solution.

Excess Electrons

Excess electrons are additional electrons that create a net negative charge on an object. Determining the number of excess electrons involves calculating the net charge and then converting this charge to the number of electrons, using each electron's charge value of \(-1.6 \times 10^{-19}\) C.
After calculating the required charge to make the electric field just outside the sphere 1150 N/C, the number of electrons is given by:

• \( n = \frac{Q}{1.6 \times 10^{-19}} \)

This calculation allows us to quantify exactly how much charge, in terms of electron numbers, is necessary to produce the desired electric field around the sphere.