/*! This file is auto-generated */ .wp-block-button__link{color:#fff;background-color:#32373c;border-radius:9999px;box-shadow:none;text-decoration:none;padding:calc(.667em + 2px) calc(1.333em + 2px);font-size:1.125em}.wp-block-file__button{background:#32373c;color:#fff;text-decoration:none} Problem 84 A small sphere with mass \(m\) c... [FREE SOLUTION] | 91Ó°ÊÓ

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A small sphere with mass \(m\) carries a positive charge \(q\) and is attached to one end of a silk fiber of length \(L\). The other end of the fiber is attached to a large vertical insulating sheet that has a positive surface charge density \(\sigma\). Show that when the sphere is in equilibrium, the fiber makes an angle equal to arctan (\(q\sigma/2mg\epsilon_0\)) with the vertical sheet.

Short Answer

Expert verified
The angle \( \theta \) is \( \arctan\left(\frac{q\sigma}{2mg\epsilon_0}\right) \).

Step by step solution

01

Draw a Free-Body Diagram

First, identify all the forces acting on the sphere. The sphere experiences three forces: 1) the gravitational force downwards, \( F_g = mg \), 2) the electric force due to the charged sheet, \( F_e \), and 3) the tension, \( T \), in the fiber, which acts along the fiber.
02

Determine the Electric Force

The electric field \( E \) created by the sheet with charge density \( \sigma \) is given by the formula: \( E = \frac{\sigma}{2\epsilon_0} \). The electric force on the sphere is then \( F_e = qE = \frac{q\sigma}{2\epsilon_0} \). This force acts perpendicular to the surface of the charged sheet.
03

Analyze Forces in Equilibrium

In equilibrium, the net force in both horizontal and vertical directions is zero. Horizontally, the only force component is due to the electric force. Vertically, the tension in the wire has a component that balances the gravitational force.
04

Set up the Force Balance Equations

Write the force balance equations. Horizontally: \( T \sin(\theta) = \frac{q\sigma}{2\epsilon_0} \). Vertically: \( T \cos(\theta) = mg \). These equations ensure there is no net force in either direction.
05

Solve for the Angle

Divide the horizontal force equation by the vertical force equation: \[ \frac{T \sin(\theta)}{T \cos(\theta)} = \frac{\frac{q\sigma}{2\epsilon_0}}{mg} \]This simplifies to \( \tan(\theta) = \frac{q\sigma}{2mg\epsilon_0} \). Take the arctan of both sides to find the angle: \( \theta = \arctan\left(\frac{q\sigma}{2mg\epsilon_0}\right) \).

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Key Concepts

These are the key concepts you need to understand to accurately answer the question.

Electric field
An electric field is a zone around a charged object where other charges can feel a force. Imagine it like an invisible aura surrounding the object. It is produced by charges, and it influences other charges that enter this zone.

In our problem, the electric field is generated by a large vertical insulating sheet with a positive surface charge density \(\sigma\). The fundamental property of this field is expressed as the force per unit charge. This makes it easy to calculate how much force a charge will feel once inside this field.

For a charged plane, we use the formula:
  • \(E = \frac{\sigma}{2\epsilon_0}\)
This equation indicates that the electric field (\(E\)) from the sheet depends on the charge density (\(\sigma\)) and the permittivity of free space (\(\epsilon_0\)).
This field acts perpendicular to the sheet, providing a directional context for the problem.
Coulomb's law
Coulomb's Law helps us understand how charges interact with each other. It explains the relationship between two charged objects. The law states that like charges repel each other, while opposite charges attract. The size of this force depends on the amount of charge and the distance between them.

For our exercise, while Coulomb's Law directly calculates forces between point charges, it establishes the basis for understanding electric forces such as the one between a charged sphere and a charged sheet. Here, the electric force experienced by the sphere can be calculated as:
  • \(F_e = qE\)
Substituting the electric field (\(E\)), we get:
  • \(F_e = \frac{q\sigma}{2\epsilon_0}\)
This force acts perpendicular to the sheet and pushes the sphere away from it due to the repulsion between like charges.
Equilibrium of forces
In physics, equilibrium occurs when all forces acting on an object are balanced. This means that the net force is zero — the object either remains at rest or moves with constant velocity. In this problem, the charged sphere needs to be in equilibrium.

We organize this by looking at the forces horizontally and vertically:
  • Horizontally: The electric force (\(F_e\)) acts as the only force, since it is perpendicular to the charge sheet.
  • Vertically: Gravity acts downward, and the tension in the silk fiber counters it. The tension is in the opposite direction vertically with a component that matches the gravitational force.
To maintain this balance, the tension's horizontal component is set to the electric force:\( T \sin(\theta) = \frac{q\sigma}{2\epsilon_0} \), and the vertical component to gravity: \ (T \cos(\theta) = mg)\. These conditions prevent the sphere from moving in any direction.
Gravitational force
Gravitational force is one of the fundamental forces in nature. It attracts any two masses towards each other. In our problem, this force acts on the charged sphere, pulling it downwards towards the Earth. This is expressed simply by the formula:
  • \(F_g = mg\)
Here, \(m\) represents the mass of the sphere and \(g\) the acceleration due to gravity (approximately 9.8 m/s² on Earth).

When the sphere is hanging by the silk fiber, this gravitational force is counteracted by the vertical component of the tension in the fiber. This balance allows the sphere to remain in equilibrium, ensuring that it doesn’t fall off or rise up. Gravity thus plays a crucial role in the delicate balance of forces necessary to satisfy equilibrium conditions in our exercise.

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Most popular questions from this chapter

If a proton and an electron are released when they are 2.0 x 10\(^{-10}\) m apart (a typical atomic distance), find the initial acceleration of each particle.

A proton is placed in a uniform electric field of 2.75 \(\times 10^3 \space N/C\). Calculate (a) the magnitude of the electric force felt by the proton; (b) the proton's acceleration; (c) the proton's speed after 1.00 \(\mu\)s in the field, assuming it starts from rest.

A uniform electric field exists in the region between two oppositely charged plane parallel plates. A proton is released from rest at the surface of the positively charged plate and strikes the surface of the opposite plate, 1.60 cm distant from the first, in a time interval of 3.20 \(\times \space 10^{-6}\) s. (a) Find the magnitude of the electric field. (b) Find the speed of the proton when it strikes the negatively charged plate.

Point charge \(q_1 = -6.00 \times \space 10^{-6}\) C is on the \(x\)-axis at \(x = -0.200\space \mathrm{m}\). Point charge \(q_2\) is on the \(x\)-axis at \(x = +0.400 \space \mathrm{m}\). Point charge \(q_3 = +3.00 \times \space 10^{-6}\) C is at the origin. What is \(q_2\) (magnitude and sign) (a) if the net force on \(q_3\) is \(6.00 \mathrm{N}\) in the \(+x-\mathrm{direction}\); (b) if the net force on \(q_3\) is \(6.00 \mathrm{N}\) in the \(-x-\mathrm{direction}\)?

A charge \(q_1 = +\)5.00 nC is placed at the origin of an \(xy\)-coordinate system, and a charge \(q_2 = -\)2.00 nC is placed on the positive \(x\)-axis at \(x = \)4.00 cm. (a) If a third charge \(q_3 = +\)6.00 nC is now placed at the point \(x =\) 4.00 cm, \(y =\) 3.00 cm, find the \(x\)- and \(y\)-components of the total force exerted on this charge by the other two. (b) Find the magnitude and direction of this force.

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