Problem 81 - Fundamentals of Physics

Question

A spherical ball of charged particles has a uniform charge density. In terms of the ball's radius \(R,\) at what radial distances (a) inside and (b) outside the ball is the magnitude of the ball's electric field equal to \(\frac{1}{4}\) of the maximum magnitude of that field?

Step-by-Step Solution

Answer

Inside: \( r = \frac{R}{4} \); Outside: \( r = 2R \).

Step 1: Understand the problem

We have a sphere with a radius \( R \) and a uniform charge distribution. We need to find radial distances inside and outside the sphere where the electric field is \( \frac{1}{4} \) of its maximum value.

Step 2: Determine the maximum electric field inside the sphere

The electric field inside a uniformly charged sphere (at radius \( r < R \)) is given by Enclosed charge is proportional to \( r^3 \). So, electric field inside is proportional to \( r \).\[ E = \frac{kQr}{R^3} \]Where \( k \) is Coulomb's constant and \( Q \) is the total charge. The maximum field inside is at \( r = R \), which is \( E_{max} = \frac{kQ}{R^2} \).

Step 3: Calculate the radius inside where the field is one-fourth of the maximum

We need \( E = \frac{1}{4} E_{max} \). Substitute \( E_{max} \)\[ \frac{kQr}{R^3} = \frac{1}{4} \cdot \frac{kQ}{R^2} \]Simplifying, cancel \( kQ \),\[ \frac{r}{R^3} = \frac{1}{4R^2} \]Multiply both sides by \( R^3 \) to solve for \( r \):\[ r = \frac{R}{4} \]

Step 4: Determine electric field outside the sphere

For \( r > R \), the electric field behaves as if all charge is at the center:\[ E = \frac{kQ}{r^2} \]Maximum outside is at the surface \( r = R \), same as inside maximum at surface \( \frac{kQ}{R^2} \).

Step 5: Calculate the radius outside where the field is one-fourth of the maximum

Set the field equation to \( \frac{1}{4} E_{max} \):\[ \frac{kQ}{r^2} = \frac{1}{4} \cdot \frac{kQ}{R^2} \]Cancel \( kQ \), and solve for \( r \):\[ \frac{1}{r^2} = \frac{1}{4R^2} \]Multiplying both sides by \( r^2 \) gives\[ 4R^2 = r^2 \]Taking the square root yields\[ r = 2R \]

Step 6: Conclusion

For the electric field to be one-fourth of its maximum magnitude: inside the sphere occurs at \( r = \frac{R}{4} \), and outside the sphere occurs at \( r = 2R \).

Key Concepts

Coulomb's Law

Coulomb's Law is fundamental in understanding electric forces between charged particles. It states that the force between two point charges is directly proportional to the product of their charges and inversely proportional to the square of the distance between them. This is described mathematically as:

• The formula: \( F = k \frac{q_1q_2}{r^2} \) where:
  • \( F \) is the electrostatic force,
  • \( k \) is Coulomb's constant (approximately \( 8.99 \times 10^9 \, \text{Nm}^2/\text{C}^2 \)),
  • \( q_1 \) and \( q_2 \) are the charges,
  • \( r \) is the distance between the charges.
• The direction of the force depends on the signs of the charges.

Coulomb's Law is essential in calculating electric fields, as it allows us to determine the force exerted by charges at a distance. When dealing with distributions of charge rather than point charges, this law helps in understanding how the collective charges generate an overall field.

Charge Distribution

Charge distribution refers to how electric charge is spread over a volume or a surface. In many physics problems, dealing with uniformly distributed charge is common because it simplifies calculations. Understanding charge distribution helps to model the distribution of electric fields in space effectively.

• Uniform charge distribution means that charge is spread evenly throughout a region, such as inside a sphere or on a plane.
• For a sphere of charge:
  • The total charge \( Q \) can be spread uniformly over the volume of the sphere, contributing to an electric field.
• The concept of symmetry often simplifies electric field calculations using Gauss's Law.

In the context of a spherical charge distribution, considering the whole sphere as composed of infinitesimally small elements allows us to calculate the electric field at various points, both inside and outside the charge distribution.

Electric Field Inside a Sphere

Inside a uniformly charged sphere, the electric field increases linearly from the center to the surface. This is because the enclosed charge, which influences the electric field, grows with the cube of the radius. Thus, the electric field itself depends linearly on the radial distance.For a radial distance \( r \) inside the sphere (\( r < R \)), the electric field can be expressed as:

• \( E = \frac{kQr}{R^3} \)
• Where \( Q \) is the total charge of the sphere, \( R \) is its radius, and \( k \) is Coulomb's constant.
• The maximum electric field inside the sphere occurs at the surface \( r = R \), given by \( E_{max} = \frac{kQ}{R^2} \).

At a distance where the electric field is \( \frac{1}{4} \) of the maximum (inside), the radial distance can be calculated and is found to be \( r = \frac{R}{4} \). This shows how understanding charge distribution within a sphere helps predict electric field behavior inside.

Electric Field Outside a Sphere

For a uniformly charged sphere, when observing from outside (\( r > R \)), the electric field behaves as if all the charge were concentrated at the center. This simplification comes from symmetry and aids in predicting the field's strength at various points.The electric field at a distance \( r \) outside the sphere is:

• \( E = \frac{kQ}{r^2} \)
• This indicates the field diminishes with the square of the distance from the center.
• Just like inside the sphere, the maximum field outside is at the surface with \( E_{max} = \frac{kQ}{R^2} \).

If the required field strength is \( \frac{1}{4} \) of the maximum value, solve the equation \( \frac{kQ}{r^2} = \frac{1}{4} \frac{kQ}{R^2} \). The solution for the distance \( r \) is found as \( r = 2R \). This computation shows how the electric field's dependency on distance helps in identifying specific regions where it attains a certain magnitude.