Question

A solid conducting sphere of radius \(R_{1}=1.435 \mathrm{~m}\) has a charge of Q evenly distributed over its surface. A second solid conducting sphere of radius \(R_{2}=0.6177 \mathrm{~m}\) is initially uncharged and at a distance of \(10.00 \mathrm{~m}\) from the first sphere. The two spheres are momentarily connected with a wire, which is then removed. The resulting charge on the second sphere is \(0.9356 \mu \mathrm{C}\). What was the original charge, \(Q\), on the first sphere?

Short answer

Answer: To find the original charge Q on Sphere 1, follow these steps: 1. Calculate the surface charge density for Sphere 2 after the spheres are connected. 2. Since both spheres share the same charge density, find the charge on Sphere 1 after the connection. 3. Determine the charge lost by Sphere 1 during the process. 4. Add the charge lost by Sphere 1 to the final charge on Sphere 1 to find the original charge Q on Sphere 1.

Step-by-step solution

Calculate the surface charge density for Sphere 2

To find the surface charge density (\(\sigma_2\)) of Sphere 2 after the spheres are connected, we can use the following formula: \(\sigma_2 = \frac{Q_2}{4\pi R_2^2}\) Where: - \(Q_2 = 0.9356\, \mu\,\text{C}\) (charge on Sphere 2) - \(R_2 = 0.6177\,\text{m}\) (radius of Sphere 2) - \(\sigma_2\) (surface charge density of Sphere 2)

Calculate the surface charge density for Sphere 1

Since the spheres share the same charge density, we can say: \(\sigma_1 = \sigma_2\) Let's find the charge on Sphere 1 after the spheres are connected. Call the total charge on Sphere 1 after the connection as \(Q'_1\). The surface charge density of Sphere 1 after the connection is: \(\sigma_1 = \frac{Q'_1}{4\pi R_1^2}\) Where: - \(Q'_1\) (charge on Sphere 1 after connection) - \(R_1 = 1.435\,\text{m}\) (radius of Sphere 1) - \(\sigma_1\) (surface charge density of Sphere 1)

Determine the charge lost by Sphere 1

Since both spheres have the same surface charge density, we can write the following equation: \(\frac{Q'_1}{4\pi R_1^2}=\frac{Q_2}{4\pi R_2^2}\) Rearrange the equation for \(Q'_1\): \(Q'_1 = Q_2 \frac{R_1^2}{R_2^2}\) Substitute values: \(Q'_1 = 0.9356\, \mu\,\text{C} \frac{(1.435\,\text{m})^2}{(0.6177\,\text{m})^2}\) Calculate the value of \(Q'_1\).

Calculate the original charge on Sphere 1

Since \(Q_2\) is gained by Sphere 2 during the process, we know that \(Q_2\) is lost by Sphere 1. We can relate the initial and final charges on Sphere 1 as follows: \(Q'_1 = Q - Q_2\) Rearrange the equation for \(Q\): \(Q = Q'_1 + Q_2\) Substitute values for \(Q'_1\) and \(Q_2\) and calculate the original charge \(Q\) on Sphere 1.

Key concepts

Conducting Spheres

Conducting spheres play a crucial role in electrostatics due to their uniform charge distribution. When we say a sphere is a conductor, we imply it allows electrons to move freely on its surface. This movement of electrons continues until they are evenly spread out, resulting in a consistent electrical potential over the surface. In the context of the given problem, it's essential to understand that when two conducting spheres are connected by a wire, charges redistribute until both spheres reach the same potential. This concept is a cornerstone of electrostatic equilibrium.

It's also noteworthy that the size of the sphere influences how much charge it can distribute on its surface. Larger spheres can accommodate more charge while maintaining the same surface charge density as smaller ones. Therefore, even if two spheres start with different amounts of charge, once connected, they share the charge proportional to their surface areas. This is why in the exercise, after connecting the larger Sphere 1 with the smaller Sphere 2, charges transfer until their surface charge densities equalize.

Surface Charge Density

Surface charge density is a measure of how much electric charge resides on a unit area of a surface. Represented by the Greek letter \(\sigma\), surface charge density is critical for understanding how charges spread out across a conducting object's surface. The equation \(\sigma = \frac{Q}{4\pi R^2}\) calculates the surface charge density, where \(Q\) is the total charge and \(R\) is the radius of the sphere.

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• Importance: Surface charge density dictates the strength of the electric field just outside the surface of a conductor.
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• Uniformity: In a conducting sphere, the charge density is uniform because of the repulsive forces between like charges which cause them to spread as far apart as possible.
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• Connection with Electric Field: The surface charge density is directly proportional to the electric field produced by the charged surface.

In the exercise, we calculated the surface charge density to understand how the original charge \(Q\) on the first sphere relates to the charge on the second sphere after connection. By realizing the charge density remains constant across both spheres, we were able to deduce the initial charge distribution.

Electrostatics

Electrostatics is the field of physics that studies electric charges at rest. As apart of this, we explore interactions between charged objects, electric fields, and potentials. The principles of electrostatics explain how charges interact through electric forces and how they distribute on conductors.

An intriguing concept within electrostatics is the behavior of conductors in electrostatic equilibrium: the electric field inside a conductor is zero, leading to some fascinating outcomes, such as the surface charge distribution we've examined in the exercise. The phenomena that occur when two conductors come into contact, and the subsequent charge redistribution leading to a uniform electric potential, is a practical application of electrostatics.

In our solved problem, we applied electrostatics concepts to find the initial charge on a conducting sphere by relating it to the charge transfer between two spheres after they are momentarily connected. This application not only illustrates electrostatic equilibrium but also shows how interconnected these principles are in solving real-world problems.