Problem 28 - University Physics with Modern Physics

Question

A solid conducting sphere has net positive charge and radius \(R =\) 0.400 m. At a point 1.20 m from the center of the sphere, the electric potential due to the charge on the sphere is 24.0 V. Assume that \(V = 0\) at an infinite distance from the sphere. What is the electric potential at the center of the sphere?

Step-by-Step Solution

Answer

The electric potential at the center of the sphere is 72.0 V.

Step 1: Determine the Electric Potential Outside the Sphere

For a conducting sphere with a charge, the electric potential at a distance \(r\) from the center (where \(r > R\)) is given by the formula: \\[ V = \frac{kQ}{r} \] \where \(k\) is Coulomb's constant \(8.99 \times 10^9 \, \text{N} \, \text{m}^2/\text{C}^2\), \(Q\) is the charge on the sphere, and \(r = 1.20 \, \text{m}\). Given that \(V = 24.0 \, \text{V}\) at \(r = 1.20 \, \text{m}\), we can write: \\[ 24.0 = \frac{8.99 \times 10^9 \cdot Q}{1.20} \] \This equation allows us to solve for the charge \(Q\).

Step 2: Solve for Charge Q

Rearrange the formula from Step 1 to solve for the charge \(Q\): \\[ Q = \frac{1.20 \times 24.0}{8.99 \times 10^9} \] \Calculate the charge \(Q\).

Step 3: Determine Electric Potential Inside the Sphere

For a conducting sphere, the electric potential at any point inside or on the surface of the sphere is constant and equal to the electric potential at the surface. Therefore, we need to find the potential at \(r = R\), which is closer than \(r = 1.20 \, \text{m}\). Using the formula from Step 1: \\[ V = \frac{8.99 \times 10^9 \cdot Q}{0.400} \] \Since \(V\) inside a conductor is constant, this is also the potential at the center of the sphere.

Step 4: Calculate the Electric Potential at the Center

From Step 3, substitute the value of \(Q\) calculated in Step 2 to find the potential at \(r = 0\) (the center of the sphere). Perform the calculation: \\[ V = \frac{8.99 \times 10^9 \cdot Q}{0.400} \] \Thus, the value of \(V\) at the center of the sphere is determined.

Key Concepts

Conducting Sphere

A conducting sphere is an excellent example used to study electric potentials and fields. It is a solid object capable of transitioning an electric charge evenly over its surface. Imagine this sphere as a host that possesses the unique ability to uniformly distribute any charge it might hold.
The beauty of a conducting sphere lies in its uniformity. The sphere ensures that every part of its surface has the same electric potential. Whether we touch the sphere in one spot or another, the electrical potential stays consistent. This uniform distribution of the potential is vital for understanding how charges behave in conductive materials.
When there is an electric charge on the sphere, it affects the electric potential both on the surface and at a distance from it. Remember, the potential inside the sphere remains uniform and equal to the surface potential, a fact that holds because charges in a conductor move until they reach a balanced state.

Coulomb's Law

Coulomb's Law is a fundamental principle that describes the force between two charges. It tells us that the force of attraction or repulsion between two charged objects is directly proportional to the product of the charges and inversely proportional to the square of the distance between them. Mathematically, it is given by:
\[ F = k \frac{{|Q_1 Q_2|}}{r^2} \]
where:

• \(F\) is the force between the charges,
• \(k\) is Coulomb's constant, approximately \(8.99 \times 10^9 \, \text{N} \, \text{m}^2/\text{C}^2\),
• \(Q_1\) and \(Q_2\) are the amounts of the charges,
• \(r\) is the distance between the centers of the two charges.

Understanding Coulomb's Law is key to exploring electric forces and fields. In our problem, the law helps us calculate the electric potential caused by a charge on a conducting sphere. This principle allows us to determine how the electric potential varies with distance and aids in solving complex electrostatic problems.

Electric Charge

Electric charge is a fundamental property of particles, crucial for understanding how electrical phenomena occur. Charges are often labeled as positive or negative and are quantified in terms of coulombs \(C\).
There are key points to remember about electric charges:

• Like charges repel each other, while opposite charges attract.
• The unit of electric charge is the coulomb.
• Charges can't be created or destroyed but can be transferred from one conductor to another.

When working with conducting spheres, the electric charge is distributed along the surface. Even when an attempt to move them would occur, the electrons will try to disperse to reduce potential energy, eventually spreading evenly.
Understanding how charges work aids in calculating the electric potential due to a certain charge, which can be measured using voltage. In problems like the one we are solving, electric charge plays a critical role, as it helps establish the electric potential seen at various distances from the sphere.

Electric Potential Inside a Conductor

Electric potential inside a conductor is a fascinating concept that showcases the unique properties of conductive materials. When a conductor is in electrostatic equilibrium, its electric potential is constant throughout its entire volume, regardless of the charge it carries. This is because charges in a conductor rearrange themselves until this uniform potential is achieved, removing all electric fields from within the conductor's surface.
When dealing with a charged conducting sphere, the electric potential at any point inside the sphere is the same as at its surface. This uniformity arises because the charges reside only on the surface, leaving the interior with no uneven electric field influences.
The equation outlining the electric potential at the surface, \[ V = \frac{kQ}{R} \], also describes the potential at the center of a conducting sphere. Here, \(V\) is the electric potential, \(k\) is Coulomb's constant, \(Q\) is the charge, and \(R\) is the radius of the sphere.
Thus, once we know the potential at the surface of the sphere, we automatically know it for any point inside the sphere, simplifying calculations significantly.