Problem 67 - Fundamentals of Physics

Question

The electric field at point \(P\) just outside the outer surface of a hollow spherical conductor of inner radius \(10 \mathrm{~cm}\) and outer radius \(20 \mathrm{~cm}\) has magnitude \(450 \mathrm{~N} / \mathrm{C}\) and is directed outward. When a particle of unknown charge \(Q\) is introduced into the center of the sphere, the electric field at \(P\) is still directed outward but is now \(180 \mathrm{~N} / \mathrm{C}\). (a) What was the net charge enclosed by the outer surface before \(Q\) was introduced? (b) What is charge \(Q\) ? After \(Q\) is introduced, what is the charge on the (c) inner and (d) outer surface of the conductor?

Step-by-Step Solution

Answer

(a) \(2 \times 10^{-9} \, \mathrm{C}\), (b) \(-1.2 \times 10^{-9} \, \mathrm{C}\), (c) \(1.2 \times 10^{-9} \, \mathrm{C}\), (d) \(0.8 \times 10^{-9} \, \mathrm{C}\).

Step 1: Identify Initial Conditions

Initially, the electric field at point \(P\), just outside the outer surface of the spherical conductor, is \(450 \, \mathrm{N/C}\). Since the field is directed outward, it indicates a positive net charge on the sphere.

Step 2: Calculate Initial Net Charge Enclosed

The electric field just outside a conductor is related to the surface charge by the equation \(E = \frac{kQ}{r^2}\), where \(E\) is the electric field, \(Q\) is the charge, and \(r\) is the distance from the charge center. Here, \(k = 8.99 \times 10^9 \, \mathrm{N\, m^2/C^2}\) and \(r = 0.2 \, \mathrm{m}\). So, initially, the net charge is \(Q = \frac{E r^2}{k} = \frac{450 \times (0.2)^2}{8.99 \times 10^9} \approx 2 \times 10^{-9} \, \mathrm{C}\)."

Step 3: Identify Conditions After Introducing Charge Q

After introducing the charge \(Q\) to the center, the electric field at \(P\) changes to \(180 \, \mathrm{N/C}\) but is still directed outward, indicating the net charge on the outer surface has changed.

Step 4: Calculate New Net Charge Enclosed

The new electric field equation becomes \(E_{new} = \frac{k(Q + Q_{initial})}{r^2}\). Substitute the values \(E_{new} = 180 \, \mathrm{N/C}\) and solve for \(Q + Q_{initial}\) to find the new total charge. \(Q + Q_{initial} = \frac{180 \times (0.2)^2}{8.99 \times 10^9} \approx 0.8 \times 10^{-9} \, \mathrm{C}\)."

Step 5: Determine Charge Q

Using the difference in net charges before and after introducing \(Q\), we have \(Q_{effective} = Q_{initial} + Q = 0.8 \times 10^{-9} \, \mathrm{C}\), so \(Q = 0.8 \times 10^{-9} - 2 \times 10^{-9} = -1.2 \times 10^{-9} \, \mathrm{C}\)."

Step 6: Determine Charges on Inner and Outer Surfaces

After \(Q\) is introduced, the inner surface becomes induced with a charge equal in magnitude but opposite to \(Q\), hence \(-(-1.2 \times 10^{-9} \, \mathrm{C}) = 1.2 \times 10^{-9} \, \mathrm{C}\) on the inner surface. The outer surface carries the remaining net charge \(2 \times 10^{-9} \, \mathrm{C} - 1.2 \times 10^{-9} \, \mathrm{C} = 0.8 \times 10^{-9} \, \mathrm{C}\)."

Key Concepts

Electric field

When we think about electric fields, it's like thinking about invisible lines pushing or pulling charged particles. The strength of this push or pull at any point is what we refer to as the electric field. The electric field is a vector, which means it has both a direction and a magnitude (or strength), described as \( E \) and measured in \( ext{N/C}\) (newtons per coulomb).
A positive charge creates an outward field and a negative charge creates an inward field. When we say the electric field at point \( P \) is \( 450 \, ext{N/C} \/ \, 180 \, ext{N/C} \,\) directed outward, it tells us there is a net positive charge affecting that area.

• The field tells us how a test charge would move if placed at that point.
• Calculations use \( E = \frac{kQ}{r^2} \), where \( Q \) is the charge and \( r \) is the distance from the charge.
• The constant \( k \), or Coulomb’s constant, has a value of \( 8.99 \, \times \, 10^9 \, ext{N} \cdot \, ext{m}^2/ ext{C}^2 \).

Surface charge

The concept of surface charge helps us understand how charges reside on conductors. In a conductor, static charges move to the surface, spreading out to minimize repulsion between like charges. This spreading is known as the surface charge density, or how much charge resides per unit area.
For a spherical conductor, the surface charge determines the electric field just outside the surface.

• The initial surface charge causes an outward electric field of \( 450 \, ext{N/C} \, \).
• After introducing a charge \( Q \), you observe changes in the surface charge at \( 180 \, ext{N/C} \, \).
• This change reflects the additional or reduced total charge affecting the outer surface.

Understanding surface charge helps determine how much charge resides at different parts of a conductor after modifications, as charges spread out in response to changes.

Gauss's Law

Gauss's Law provides a powerful tool for understanding electric fields and charges. It states that the total electric flux through a closed surface is equal to the charge enclosed divided by the permittivity of free space. Mathematically, it's written as \( \, ext{Φ} = \frac{Q_{ ext{enc}}}{ ext{ε}_0} \).
This principle helps solve complex problems involving symmetric charge distributions, like the hollow spherical conductor in our exercise.

• Before introducing \( Q \, \), the sphere encloses a net charge that determines the initial electric field of \( 450 \, ext{N/C} \, \).
• After \( Q \, \), it modifies the charge distribution while maintaining radial symmetry, keeping field calculations straightforward due to the sphere's symmetry.
• The law simplifies calculating induced and residual charges by using symmetry and charge conservation.

This approach enables accurate calculations of electric fields generated by different charge configurations.

Electric potential

Electric potential gives insight into the energy profile of electric fields. It is a measure of the work done in bringing a charge from infinity to a point in space, without acceleration. Electric potential is like the potential energy in a gravitational field, often described in volts (V). It relates closely to the electric field since the field can be seen as a gradient of electric potential.
The electric potential difference between two points in an electric field tells us the work done in moving a unit charge between these points.

• Initial and post-modification fields at \( P \) represent certain electric potentials due to respective surface charges.
• The change due to charge \( Q \) affects the electric potential, causing a difference in work done to bring a charge to \( P \).
• This is critical in understanding energy changes when charges redistribute in conductors.

Recognizing changes in potential allows us to understand how introducing net charges affects both surface fields and potential.