Problem 83 - University Physics with Modern Physics

Question

A metal sphere with radius \(R_{1}\) has a charge \(Q_{1}\) . Take the electric potential to be zero at an infinite distance from the sphere. (a) What are the electric field and electric potential at the surface of the sphere? This sphere is now connected by a long, thin conduct- ing wire to another sphere of radius \(R_{2}\) that is several meters from the first sphere. Before the connection is made, this second sphere is uncharged. After electrostatic equilibrium has been reached, what are (b) the total charge on each sphere; (c) the electric potential at the surface of each sphere; (d) the electric field at the surface of each sphere? Assume that the amount of charge on the wire is much less than the charge on each sphere.

Step-by-Step Solution

Answer

(a) \(E = \frac{kQ_1}{R_1^2}\), \(V = \frac{kQ_1}{R_1}\). (b) \(Q_1' = \frac{Q_1 R_2}{R_1 + R_2}\), \(Q_2' = \frac{Q_1 R_1}{R_1 + R_2}\). (c) \(V = \frac{kQ_1}{R_1 + R_2}\). (d) \(E_1 = \frac{kQ_1'R_2}{R_1(R_1 + R_2)}\), \(E_2 = \frac{kQ_1'R_1}{R_2(R_1 + R_2)}\).

Step 1: Calculate Electric Field at Sphere Surface

The electric field \(E\) just outside the surface of a charged sphere of radius \(R_1\) with charge \(Q_1\) is given by Gauss's law as\[E = \frac{kQ_1}{R_1^2}\]where \(k\) is the electrostatic constant \(8.99 \times 10^9 \; \text{Nm}^2/\text{C}^2\). The field is radially outward if \(Q_1\) is positive, and inward if \(Q_1\) is negative.

Step 2: Calculate Electric Potential at Sphere Surface

The electric potential \(V\) at the surface of a charged sphere is given by\[V = \frac{kQ_1}{R_1}\]This is because the potential is set to zero at infinity and only depends on the charge and radius.

Step 3: Determine Charges After Connecting Two Spheres

When the spheres are connected by a conducting wire, charges redistribute such that both spheres have the same electric potential \(V\). If the total charge after redistribution is still \(Q_1\) (since the second sphere is initially uncharged), then the charge on each sphere \(Q_1'\) and \(Q_2'\) must satisfy:\[V = \frac{kQ_1'}{R_1} = \frac{kQ_2'}{R_2} \quad \text{and} \quad Q_1' + Q_2' = Q_1\]Solving these equations gives the redistributed charges on each sphere:\[Q_1' = \frac{Q_1 R_2}{R_1 + R_2}, \quad Q_2' = \frac{Q_1 R_1}{R_1 + R_2}\]

Step 4: Calculate Electric Potential at Surface of Each Sphere

After equilibrium, both spheres have the same potential given by\[V = \frac{kQ_1'}{R_1} = \frac{kQ_2'}{R_2}\]Thus, substitute for \(Q_1'\) or \(Q_2'\) to find the common potential\[V = \frac{kQ_1}{R_1 + R_2}\]

Step 5: Calculate Electric Field at Surface of Each Sphere

The electric field at the surface of each sphere after charge redistribution can be calculated using\[E_1 = \frac{kQ_1'}{R_1^2} = \frac{k \times \left(\frac{Q_1 R_2}{R_1 + R_2}\right)}{R_1^2}\]\[E_2 = \frac{kQ_2'}{R_2^2} = \frac{k \times \left(\frac{Q_1 R_1}{R_1 + R_2}\right)}{R_2^2}\]These are the electric fields at the surfaces of spheres 1 and 2 respectively.

Key Concepts

Electric Field

The electric field is a crucial principle in electrostatics, representing the force per unit charge that a charge experiences due to another charge. With a sphere of radius \(R_1\) and charge \(Q_1\), Gauss's law helps in calculating this field. According to Gauss's law,

• The electric field \(E\) just outside a charged sphere ensures it is radially outward or inward depending on the charge being positive or negative.
• This field strength is mathematically given by:
  \[ E = \frac{kQ_1}{R_1^2} \]
  where \(k\) is the electrostatic constant \(8.99 \times 10^9 \, \text{Nm}^2/\text{C}^2\). This formula derives from the symmetry of the sphere.

By understanding these calculations, one can predict how charges will act near such a sphere.

Electric Potential

Electric potential is another core concept that describes the work needed to move a charge to a particular point in the field. This potential is zero at an infinite distance from the charge. For a sphere with charge \(Q_1\) and radius \(R_1\), the potential \(V\) at its surface can be found by:

• Using the formula:
  \[ V = \frac{kQ_1}{R_1} \]
• Electric potential is scalar and accumulates from point to point, making it simple to compute for spherical cases.

This potential explains how much work is needed to bring a unit charge from infinity to the sphere's surface.

Gauss's Law

Gauss's law is essential in calculating the electric fields for symmetrical charge distributions. It states that the total electric flux through a closed surface equals the charge enclosed divided by the electric constant. In this scenario:

• Gauss’s law simplifies calculations for spherical geometries by allowing the use of symmetry considerations.
• The equation
  \[ E = \frac{kQ_1}{R_1^2} \]
  emerges from integrating the electric field over the sphere's surface.

This law is powerful for solving problems involving electric fields in various symmetrical setups. It aids in understanding how charge distributions affect the field around a conductor or insulator.

Electrostatic Equilibrium

When dealing with electric charges and conductors, understanding electrostatic equilibrium is crucial. After connecting two conducting spheres, charges redistribute until both have the same electrostatic potential. Important aspects include:

• Charge redistribution ensures both spheres reach the same potential \(V\). This is calculated as:
  \[ V = \frac{kQ_1}{R_1 + R_2} \]
• At equilibrium, resulting charges \(Q_1'\) and \(Q_2'\) satisfy the potential balance. They can be calculated using:
  \[ Q_1' = \frac{Q_1 R_2}{R_1 + R_2}, \quad Q_2' = \frac{Q_1 R_1}{R_1 + R_2} \]

Electrostatic equilibrium helps predict the final state of charge distribution when conductive materials interact. This concept is vital in designing and understanding systems involving multiple charged bodies or shells.