Problem 44 - University Physics with Modern Physics

Question

The electric field at the surface of a charged, solid, copper 0sphere with radius 0.200 \(\mathrm{m}\) is 3800 \(\mathrm{N} / \mathrm{C}\) , directed toward the center of the sphere. What is the potential at the center of the sphere, if we take the potential to be zero infinitely far from the sphere?

Step-by-Step Solution

Answer

The potential at the center is 760 V.

Step 1: Identify Known Quantities

We know the radius of the sphere, \( r = 0.200 \, \mathrm{m} \), and the electric field at the surface, \( E = 3800 \, \mathrm{N/C} \). The electric field is directed toward the center indicating a negative charge.

Step 2: Use Gauss's Law to Find Charge

Using Gauss's law, the electric field due to a charged sphere is given by \( E = \frac{kQ}{r^2} \), where \( k \) is Coulomb's constant \( 8.99 \times 10^9 \, \mathrm{N} \, \mathrm{m}^2/\mathrm{C}^2 \) and \( Q \) is the charge. Solving for \( Q \), we have \( Q = \frac{E r^2}{k} \). Substituting the values, \( Q = \frac{3800 \times (0.200)^2}{8.99 \times 10^9} \), which gives \( Q \approx 1.69 \times 10^{-8} \, \mathrm{C} \), indicating negative charge.

Step 3: Calculate Potential at the Surface

The potential \( V \) at the surface of the sphere is given by \( V = \frac{kQ}{r} \). Substituting for \( Q \), \( V = \frac{8.99 \times 10^9 \times 1.69 \times 10^{-8}}{0.200} \approx 760 \, \mathrm{V} \).

Step 4: Determine Potential at the Center

For a conductor in electrostatic equilibrium, the electric potential is constant throughout the conductor. Therefore, the potential at the center of the sphere is the same as at the surface, which is 760 V.

Key Concepts

Gauss's Law

Gauss's Law is a powerful tool used to calculate electric fields, particularly in situations with great symmetry. It relates an electric field around a closed surface to the charge enclosed by that surface. The mathematical formulation of Gauss's Law is \( \oint \mathbf{E} \cdot d\mathbf{A} = \frac{Q_{enc}}{\varepsilon_0} \), where \( \oint \mathbf{E} \cdot d\mathbf{A} \) is the electric flux through a closed surface, \( Q_{enc} \) is the charge enclosed within that surface, and \( \varepsilon_0 \) is the permittivity of free space.

• Imagine surrounding a charged object with an imaginary surface called a "Gaussian surface."
• If the electric field is uniform, as in our exercise's sphere, it simplifies to \( E \cdot A = \frac{Q}{\varepsilon_0} \), where \( A \) is the area of the sphere.
• For a sphere, the surface area is \( 4\pi r^2 \), which allows us to rearrange the formula to solve for charge \( Q \), as done in our solution.

Using this law, we determined the charge on the sphere, useful for finding other electrical properties like the potential.

Electric Field

The electric field represents the force per unit charge exerted on a small positive test charge placed in the field. It is a vector quantity with both magnitude and direction. For a charged sphere, the electric field \( E \) at its surface is given by \( E = \frac{kQ}{r^2} \), where \( k \) is Coulomb's constant, \( Q \) is the charge on the sphere, and \( r \) is the radius.
Here are a few points about electric fields:

• The direction of the electric field shows the direction of force that would act on a positive charge. In our exercise, the field is directed inward, indicating a negative charge.
• Magnitudes reflect field strength, which can be calculated using measured or known quantities.
• In our solid sphere exercise, we used the surface electric field information to derive the charge \( Q \) via Gauss’s Law.

Understanding electric fields helps us predict how charges will interact and move within different spaces.

Electric Potential

Electric potential, often referred to as voltage, is the potential energy per unit charge at a point in an electric field. It tells us how much work is needed to move a charge within the field. For a spherical charge distribution, the potential \( V \) at its surface is \( V = \frac{kQ}{r} \).
Several key points regarding electric potential are:

• In our exercise, we calculated the potential at the sphere's surface using its charge and radius.
• Electric potential decreases as you move further from the charge, approaching zero infinitely far away, which is a common reference point.
• The potential is evenly distributed throughout a conductor in electrostatic equilibrium, meaning the potential inside the sphere remains the same as on its surface.

Understanding electric potential is crucial when considering how charges are influenced and move in an electric field. It helps in visualizing energy landscapes in electrodynamics.

Conductors in Electrostatic Equilibrium

Conductors in electrostatic equilibrium have special properties due to the redistribution of charges. When a conductor reaches equilibrium, the following are true:

• The electric field inside the conductor is zero. This occurs because any internal electric fields would cause charges to move, disturbing equilibrium.
• The charge resides entirely on the surface of the conductor, ensuring no net field within.
• The potential throughout the conductor is constant. This is why, in our exercise, the electric potential at the center of the sphere equals that at the surface. Charges in conductors move until a uniform potential is achieved.

These characteristics ensure that the conductor conducts electricity efficiently, redirecting or concentrating charge where necessary. Understanding conductors under electrostatic conditions is essential for grasping how materials behave in electric fields.