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At time \(t =\) 0 a proton is a distance of 0.360 m from a very large insulating sheet of charge and is moving parallel to the sheet with speed 9.70 \(\times\) 10\(^2\) m/s. The sheet has uniform surface charge density 2.34 \(\times\) 10\(^{-9}\) C/m2. What is the speed of the proton at \(t =\) 5.00 \(\times\) 10\(^{-8}\) s?

Short Answer

Expert verified
The speed of the proton at \( t = 5.00 \times 10^{-8} \text{ s} \) is approximately 1603.5 m/s.

Step by step solution

01

Understand the Electric Field

We start by calculating the electric field due to the very large insulating sheet with uniform surface charge density \( \sigma = 2.34 \times 10^{-9} \) C/m². The electric field \( E \) from an infinite plane with charge density \( \sigma \) is given by the formula \( E = \frac{\sigma}{2\varepsilon_0} \), where \( \varepsilon_0 \) is the vacuum permittivity, approximately \( 8.85 \times 10^{-12} \text{ C}^2/\text{N}\cdot\text{m}^2 \).
02

Calculate the Electric Field

Calculate the electric field \( E \) using the formula from Step 1: \[ E = \frac{2.34 \times 10^{-9}}{2 \times 8.85 \times 10^{-12}} = \frac{2.34 \times 10^{-9}}{1.77 \times 10^{-11}} = 132.2 \text{ N/C} \].
03

Determine Force on the Proton

The force \( F \) acting on the proton can be calculated using \( F = qE \), where \( q = 1.6 \times 10^{-19} \) C is the charge of a proton. Therefore, \( F = 1.6 \times 10^{-19} \times 132.2 = 2.1152 \times 10^{-17} \text{ N} \).
04

Calculate the Acceleration of the Proton

We use Newton's second law to find the acceleration \( a \) of the proton. Newton's second law states \( F = ma \), where \( m = 1.67 \times 10^{-27} \) kg is the mass of the proton. Solve for \( a \): \[ a = \frac{F}{m} = \frac{2.1152 \times 10^{-17}}{1.67 \times 10^{-27}} = 1.267 \times 10^{10} \text{ m/s}^2 \].
05

Calculate Final Velocity

Using the kinematic equation \( v = u + at \), where \( u = 9.70 \times 10^2 \text{ m/s} \) is the initial speed of the proton, \( a = 1.267 \times 10^{10} \text{ m/s}^2 \) is the acceleration, and \( t = 5.00 \times 10^{-8} \) s is the time: \[ v = 9.70 \times 10^2 + 1.267 \times 10^{10} \times 5.00 \times 10^{-8} \].
06

Evaluate the Final Velocity

Perform the calculation: \[ v = 970 + 1.267 \times 10^{10} \times 5 \times 10^{-8} = 970 + 633.5 = 1603.5 \text{ m/s} \]. Thus, after \( 5.00 \times 10^{-8} \) seconds, the speed of the proton is approximately \( 1603.5 \text{ m/s} \).

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Key Concepts

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

Uniform Surface Charge Density
When discussing electric fields, particularly in the context of charged surfaces, it's important to understand the concept of **uniform surface charge density**. This term refers to the distribution of electric charge across a surface. A uniform surface charge density indicates that the charge is spread evenly over the surface, which simplifies the calculations for electric fields.

The electric field created by a uniformly charged infinite sheet is particularly notable. For an infinitely large sheet, the electric field is calculated as:
  • Independent of distance from the sheet
  • Uniform and constant across space
These characteristics result because the field lines are parallel and uniformly distributed. The formula to calculate this constant electric field **E** for an infinite sheet with surface charge density \(\sigma\) is:\[ E = \frac{\sigma}{2 \varepsilon_0} \]where \(\varepsilon_0\) represents the vacuum permittivity, a constant value.

This consistent electric field plays a crucial role in determining how a proton or any charged particle will behave when introduced into this field.
Proton Acceleration
Acceleration occurs when a force acts on a mass. In the context of charged particles, like a proton, moving in an electric field, this force can lead to changes in motion. **Proton acceleration** is determined by the electric force derived from the field.

The force **F** experienced by a proton in an electric field **E** is computed using the relation:\[ F = qE \]where **q** is the charge of the proton, equal to \(1.6 \times 10^{-19} \text{ C}\). Once we have the force, we apply Newton's second law, \( F = ma \), where **m** is the proton's mass \(1.67 \times 10^{-27} \text{ kg}\), to find the acceleration **a**:\[ a = \frac{F}{m} \]This allows us to calculate how fast the proton speeds up in response to the constant push it receives from the electric field.

Understanding this acceleration is critical because it tells us how the proton's velocity changes over time while it moves in the field.
Kinematic Equation
In physics, kinematics deals with the motion of objects without discussing the forces that cause such motion. The **kinematic equation** used here relates the initial velocity, acceleration, and time to determine the final velocity of an object.

Given an initial velocity **u**, an object's velocity **v** after time **t** can be calculated if it undergoes constant acceleration **a**. The kinematic equation is expressed as:\[ v = u + at \]This equation allows us to predict the final velocity of the proton as it moves through the electric field.
  • **u**: Initial speed of the proton
  • **a**: Acceleration calculated from the electric force
  • **t**: Time duration in motion
Using this equation, we can conclude that the proton's speed will increase linearly over the stated time interval if the acceleration remains constant. This is crucial for calculating how the proton's speed changes as a result of the uniform electric field, enabling us to find its new speed after **5.00 \(\times\) 10\(^{-8}\)** seconds.

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Most popular questions from this chapter

An electron is released from rest at a distance of 0.300 m from a large insulating sheet of charge that has uniform surface charge density +2.90 \(\times\) 10\(^{-12}\) C/m2. (a) How much work is done on the electron by the electric field of the sheet as the electron moves from its initial position to a point 0.050 m from the sheet? (b) What is the speed of the electron when it is 0.050 m from the sheet?

A very large, horizontal, nonconducting sheet of charge has uniform charge per unit area \(\sigma =\) 5.00 \(\times\) 10\(^{-6}\) C/m\(^2\). (a) 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. (a) If the sphere is to remain motionless when it is released, what must be the value of \(q\)? (b) What is \(q\) if the sphere is released 1.50 cm above the sheet?

The electric field 0.400 m from a very long uniform line of charge is 840 N/C. How much charge is contained in a 2.00-cm section of the line?

A very long uniform line of charge has charge per unit length 4.80 \(\mu\)C/m and lies along the \(x\)-axis. A second long uniform line of charge has charge per unit length -2.40 \(\mu\)C/m and is parallel to the x-axis at \(y =\) 0.400 m. What is the net electric field (magnitude and direction) at the following points on the \(y\)-axis: (a) \(y =\) 0.200 m and (b) \(y =\) 0.600 m?

(a) How many excess electrons must be distributed uniformly within the volume of an isolated plastic sphere 30.0 cm in diameter to produce an electric field of magnitude 1390 N/C just outside the surface of the sphere? (b) What is the electric field at a point 10.0 cm outside the surface of the sphere?

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