The image illustrates an experimental setup involving a charged particle in an electric field. Here is a detailed description of the setup:1. **Components of the Diagram**: - A charged particle, labeled with charge \( q \) and mass \( m \), enters an electric field from the left. - The particle has an initial velocity denoted by \( v_0 \). - The electric field \( \mathbf{E} \) is represented by vertical arrows pointing downward, indicating the direction of the field. - The distance covered by the electric field is labeled as 26.0 cm.2. **Path and Measurement**: - After passing through the electric field, the particle travels a further horizontal distance of 56.0 cm, presumably unaffected by the field now. - A screen is positioned at the end of this path. - There is a vertical displacement labeled \( d \) on the screen, indicating how far the particle has moved from its original path due to the action of the electric field.3. **Purpose**: - This setup is typically used to study the deflection of charged particles in an electric field, measuring their displacement \( d \) on the screen to calculate or confirm properties such as charge-to-mass ratio, velocity changes, or the characteristics of the electric field.This model is often utilized in physics experiments related to electromagnetism, helping illustrate concepts like the force exerted on a charge by an electric field and the effect of this force on the particle's trajectory. ### Physics Problem: Charge-to-Mass Ratio Calculation**Problem Description:**A small object with mass \( m \), charge \( q \), and initial speed \( v_0 = 6.00 \times 10^3 \, \text{m/s} \) is projected into a uniform electric field between two parallel metal plates of length \( 26.0 \, \text{cm} \) (refer to Figure 1). The electric field between the plates is directed downward with a magnitude \( E = 800 \, \text{N/C} \). The field is assumed to be zero outside the region between the plates.- The separation between the plates is large enough for the object to pass without hitting the lower plate.- After traversing the field region, the object is deflected downward by a vertical distance \( d = 1.35 \, \text{cm} \) from its initial direction.- A collecting plate is situated \( 56.0 \, \text{cm} \) from the edge of the parallel plates.Assumptions: Ignore the effects of gravity and air resistance.**Part A: Calculation Task**Calculate the object's charge-to-mass ratio, \( \frac{q}{m} \).- Express the answer in coulombs per kilogram (C/kg).**Diagram Explanation:**No explicit diagrams are provided in the text, but Figure 1 likely illustrates the setup, showing:- Two parallel metal plates with a uniform downward electric field \( E \).- The trajectory of the object, marking its deflection \( d \) after passing through the plates.- The position of the collecting plate relative to the plates. To solve the problem, apply concepts of electric fields, motion under uniform acceleration (deflection calculation), and kinematics.

The expression for deflection of a first object in electric field is, d1 = ut1 + 1/2a (t1)2 u=0 is…

A small object with mass m, charge q, and initial speed vo = 6.00x103 m/s is projected into a uniform electric field between two parallel metal plates of length 26.0 cm (Figure 1). The electric field between the plates is directed downward and has magnitude E = 800 N/C. Assume that the field is zero outside the region between the plates. The separation between the plates is large enough for the object to pass between the plates without hitting the lower plate. After passing through the field region, the object is deflected downward a vertical distance d = 1.35 cm from its original direction of motion and reaches a collecting plate that is 56.0 cm from the edge of the parallel plates. Ignore gravity and air resistance. Part A Calculate the object's charge-to-mass ratio, q/m. Express your answer in coulombs per kilogran q/m =

A small object with mass m, charge q, and initial speed vo = 6.00x103 m/s is projected into a uniform electric field between two parallel metal plates of length 26.0 cm (Figure 1). The electric field between the plates is directed downward and has magnitude E = 800 N/C. Assume that the field is zero outside the region between the plates. The separation between the plates is large enough for the object to pass between the plates without hitting the lower plate. After passing through the field region, the object is deflected downward a vertical distance d = 1.35 cm from its original direction of motion and reaches a collecting plate that is 56.0 cm from the edge of the parallel plates. Ignore gravity and air resistance. Part A Calculate the object's charge-to-mass ratio, q/m. Express your answer in coulombs per kilogran q/m =

College Physics

11th Edition

ISBN:9781305952300

Author:Raymond A. Serway, Chris Vuille

Publisher:Raymond A. Serway, Chris Vuille

Chapter1: Units, Trigonometry. And Vectors

Section: Chapter Questions

Problem 1CQ: Estimate the order of magnitude of the length, in meters, of each of the following; (a) a mouse, (b)...

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