A very large, horizontal, nonconducting sheet of charge has uniform charge per unit area σ = 5.00 × 10 − 6 \sigma=5.00\times10^{-6} σ = 5.00 × 1 0 − 6 C/m 2 . A small sphere of mass m = 8.00 × 10 − 6 m=8.00\times10^{-6} m = 8.00 × 1 0 − 6 kg and charge q q is placed 3.00 3.00 3.00 cm above the sheet of charge and then released from rest. What is q q if the sphere is released 1.50 1.50 cm above the sheet?

Hey everyone. So this problem is dealing with charged particles. Let's see what they are giving us and what they're asking from us. So we know we have a particle with a certain mass and a given charge, it's at rest, this given height five centimeters above a large flat nonconducting plate. That plate has a constant charge density given to us. And they're asking us the charge of another particle where the mass is the same as the first particle that also remains stationary when placed without initial speed. So at rest uh at a distance of two centimeters above the plate. So the first thing that we need to do here is draw our free body diagram. So we have this charge and we have the electric force from the charge density acting in an upward direction. And then we have the weight of the um charge particle acting in a downward direction. So from here, let's recall that the electric force is given by QE where Q is a charge and E is the magnitude of the electric field and weight is given as mass times gravity. So we're going to solve the free body diagram for both of the um particles. So P one and P two and then um see if that tells us what the charge of the second particle is. So for particle one, we have Q one E one equals M one G gravity is a constant. So we can rewrite this as Q one equals M one G over E one. For the second particle, the free body diagram looks exactly the same. The only difference is between these two is the height away from the or above the electric field. And so Q two, it's going to be, or for particle two, it's gonna be Q two B two equals M two G where QE two equals M two G over E two. And so we are already given in the problem that M two equals M one. And we know that the um plate has a un a constant charge density, which means it has a uniform electric field. So E two equals E one. So we can rewrite this as Q one equals M one G over E one and Q two equals M one G over E one. And so from here, we can see that Q one and Q two are the same. And so therefore, Q two must equal Q one which was given to us in the problem as 2.83 times 10, the negative 11 pull ups. And so that is the answer to this problem when we look at our potential solutions. And we see that, that aligns with choice B. So that's all we have for this one, we'll see you in the next video.

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24. Electric Force & Field; Gauss' Law

Gauss' Law

Physics

24. Electric Force & Field; Gauss' Law

Gauss' Law

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Problem 26b

Textbook Question

A very large, horizontal, nonconducting sheet of charge has uniform charge per unit area $\sigma=5.00\times10^{-6}$ C/m². A small sphere of mass $m=8.00\times10^{-6}$ kg and charge $q$ is placed $3.00$ cm above the sheet of charge and then released from rest. What is $q$ if the sphere is released $1.50$ cm above the sheet?

Verified step by step guidance

Step 1: Understand the problem setup. We have a large sheet with a uniform charge density $ \sigma = 5.00 \times 10^{-6} \text{ C/m}^2 $. A small sphere with mass $ m = 8.00 \times 10^{-6} \text{ kg} $ and charge $ q $ is placed above the sheet. We need to find the charge $ q $ when the sphere is released from different heights above the sheet.

Step 2: Use Gauss's Law to find the electric field $ E $ due to the sheet. For an infinite sheet of charge, the electric field is given by $ E = \frac{\sigma}{2\varepsilon_0} $, where $ \varepsilon_0 $ is the permittivity of free space $ \varepsilon_0 = 8.85 \times 10^{-12} \text{ C}^2/\text{N} \cdot \text{m}^2 $.

Step 3: Calculate the force $ F $ acting on the sphere due to the electric field. The force is given by $ F = qE $. Since the sphere is released from rest, this force will cause it to accelerate towards the sheet.

Step 4: Apply Newton's second law to relate the force to the acceleration $ a $ of the sphere. According to Newton's second law, $ F = ma $, where $ m $ is the mass of the sphere. Therefore, $ qE = ma $.

Step 5: Solve for the charge $ q $. Rearrange the equation $ qE = ma $ to find $ q = \frac{ma}{E} $. Substitute the values for $ m $, $ a $, and $ E $ to find $ q $. Note that the acceleration $ a $ can be determined from the change in height and the initial conditions of the sphere's motion.

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Key Concepts

Here are the essential concepts you must grasp in order to answer the question correctly.

Electric Field due to a Charged Sheet

The electric field generated by an infinite sheet of charge is uniform and perpendicular to the surface. It is given by E = σ / (2ε₀), where σ is the charge density and ε₀ is the permittivity of free space. This field does not depend on the distance from the sheet, making it crucial for understanding the forces acting on nearby charged objects.

Force on a Charged Particle in an Electric Field

A charged particle in an electric field experiences a force given by F = qE, where q is the charge of the particle and E is the electric field strength. This force influences the motion of the particle, and understanding it is essential for predicting how the sphere will move when released above the charged sheet.

Gravitational Force

The gravitational force acting on an object is given by F = mg, where m is the mass of the object and g is the acceleration due to gravity. In this scenario, the gravitational force opposes the electric force, and analyzing the balance between these forces is key to determining the charge q required for equilibrium or specific motion.

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Textbook Question

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Textbook Question

A very large, horizontal, nonconducting sheet of charge has uniform charge per unit area $σ = 5.00 \times 10^{- 6}$ C/m². A small sphere of mass $m = 8.00 \times 10^{- 6}$ kg and charge $q$ is placed $3.00$ cm above the sheet of charge and then released from rest. If the sphere is to remain motionless when it is released, what must be the value of $q$ ?

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Textbook Question

An infinitely long cylindrical conductor has radius $r$ and uniform surface charge density $σ$ . In terms of $σ$ , what is the magnitude of the electric field produced by the charged cylinder at a distance $r > R$ from its axis? Then, express the result in terms of $λ$ and show that the electric field outside the cylinder is the same as if all the charge were on the axis.

An infinitely long cylindrical conductor has radius $r$ and uniform surface charge density $σ$ . In terms of $σ$ and $R$ , what is the charge per unit length $λ$ for the cylinder?

A very long conducting tube (hollow cylinder) has inner radius $A$ and outer radius $b$ . It carries charge per unit length $+α$ , where $α$ is a positive constant with units of C/m. A line of charge lies along the axis of the tube. The line of charge has charge per unit length. Calculate the electric field in terms of $α$ and the distance $r$ from the axis of the tube for (i) $r < a$ ; (ii) $a < r < b$ ; (iii) $r > b$ . Show your results in a graph of $E$ as a function of $R$ .

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