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At \(t=0\) a very small object with mass \(0.400 \mathrm{mg}\) and charge \(+9.00 \mu \mathrm{C}\) is traveling at \(125 \mathrm{~m} / \mathrm{s}\) in the \(-x\)-direction. The charge is moving in a uniform electric field that is in the \(+y\)-direction and that has magnitude \(E=895 \mathrm{~N} / \mathrm{C}\). The gravitational force on the particle can be neglected. How far is the particle from the origin at \(t=7.00 \mathrm{~ms} ?\)

Short Answer

Expert verified
The particle is 0.875 m from the origin at \(t=7.00 \text{ ms}\).

Step by step solution

01

Identify Known Variables

First, we identify all given variables in the problem. The mass of the object is \(0.400 \text{ mg} = 0.400 \times 10^{-3} \text{ kg}\), the charge is \(+9.00 \mu \text{C} = 9.00 \times 10^{-6} \text{ C}\), the initial velocity \(v_{x0} = 125 \text{ m/s}\), the electric field \(E = 895 \text{ N/C}\), and the time \(t = 7.00 \text{ ms} = 7.00 \times 10^{-3} \text{ s}\).
02

Calculate Electric Force

The electric force \(F\) acting on the charged particle can be calculated using the formula \(F = qE\), where \(q\) is the charge and \(E\) is the electric field. Substituting the known values: \(F = (9.00 \times 10^{-6} \text{ C})(895 \text{ N/C}) = 8.055 \times 10^{-3} \text{ N}\). The force acts in the \(+y\)-direction.
03

Calculate Acceleration in y-direction

Using Newton's second law, \(F = ma\), solve for the acceleration \(a_y\) in the \(y\)-direction. Substituting \(m = 0.400 \times 10^{-3} \text{ kg}\) and \(F = 8.055 \times 10^{-3} \text{ N}\): \(a_y = \frac{8.055 \times 10^{-3}}{0.400 \times 10^{-3}} = 20.1375 \text{ m/s}^2\).
04

Calculate x-Position at t=7 ms

Since there is no force in the \(x\)-direction, the object continues at a constant velocity of \(v_x = 125 \text{ m/s}\). The displacement in the \(x\)-direction is \( x = v_{x}t = 125 \times 7.00 \times 10^{-3} = 0.875 \text{ m}\).
05

Calculate y-Position at t=7 ms

The initial velocity in the \(y\)-direction is \(v_{y0} = 0\) since the motion starts only in the \(x\)-direction. Use the kinematic equation for motion with constant acceleration: \(y = v_{y0}t + \frac{1}{2}a_yt^2\). Substituting the known values: \( y = 0 + \frac{1}{2} \times 20.1375 \times (7.00 \times 10^{-3})^2 = 4.927\times 10^{-4} \text{ m}\).
06

Determine Total Displacement

The total distance from the origin can be calculated using the Pythagorean theorem: \(d = \sqrt{x^2 + y^2}\). Substitute \(x = 0.875\) and \(y = 4.927\times 10^{-4}\) into the formula: \(d = \sqrt{(0.875)^2 + (4.927\times 10^{-4})^2} = 0.875 \text{ m}\) (since \(y\) is much smaller compared to \(x\)).

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Key Concepts

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

Charge Motion
Charge motion is an essential concept in physics, especially when dealing with electric fields and charged particles. In this exercise, we focus on how a charged particle moves due to its interaction with an electric field.

A charged particle experiences a force when it is placed in an electric field. This is due to the interaction between the charge and the field, following the equation:
  • Electric force: \( F = qE \)
where \( q \) is the charge, and \( E \) is the strength of the electric field. This force will influence how the particle moves in the field.

In this scenario, we have a small object with a charge of \(+9.00 \mu \mathrm{C}\) traveling in an electric field of \(895 \mathrm{~N/C}\) directed along the \( +y \) axis. This creates a force on the particle acting in the \( +y \) direction. Since gravitational effects are neglected here, only the electric force dictates the charge's motion in the \(y\)-direction.
Kinematics
Kinematics deals with the motion of objects without considering the forces causing the motion. Here, it is applied to describe the trajectory of our charged particle under the influence of an electric field.

One of the crucial elements in kinematics is determining positions under constant acceleration. The particle initially only has a velocity in the \(-x\)-direction at \(125 \mathrm{~m/s}\), and no initial velocity in the \(y\)-direction. Thanks to the electric force calculated earlier, acceleration in the \(y\)-direction can be determined using:
  • Acceleration: \( a_y = \frac{F}{m} \)
and is utilized in the kinematic equation:
  • Position: \( y = v_{y0}t + \frac{1}{2}a_yt^2 \)
Since the initial \(y\)-velocity \(v_{y0} = 0\), we focus on the change in position due to acceleration. By this logic, this approach helps compute the final displacement in the \(y\)-direction at a specific time \( t = 7 \mathrm{~ms}\).

In the \(x\)-direction, as there are no forces acting, the velocity remains constant. Thus, position in the \(x\)-direction is simply:
  • \( x = v_xt \)
Newton's Second Law
Newton's Second Law is fundamental when understanding the relationship between force and the motion it causes. The law states:
  • \( F = ma \)
Where \( F \) is the force applied to an object, \( m \) is the mass of the object, and \( a \) is the acceleration it experiences.

In our scenario, the particle's acceleration in the \(y\)-direction is determined by this principle. With a known mass and a force calculated from the electric field and the charge, the acceleration is computed. This value tells us how quickly the object's velocity changes in the \(y\)-direction.

Knowing the acceleration allows us to apply kinematic equations and predict the future motion and final position of the charge accurately, when the effect of gravitational force is considered negligible. Newton's Second Law thus bridges the link between the force from the electric field and the resultant motion of a charged particle.

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Most popular questions from this chapter

Two small spheres spaced \(20.0 \mathrm{~cm}\) apart have equal charge. How many excess electrons must be present on each sphere if the magnitude of the force of repulsion between them is \(4.57 \times 10^{-21} \mathrm{~N} ?\)

A proton is projected into a uniform electric field that points vertically upward and has magnitude \(E\). The initial velocity of the proton has a magnitude \(v_{0}\) and is directed at an angle \(\alpha\) below the horizontal. (a) Find the maximum distance \(h_{\max }\) that the proton descends vertically below its initial elevation. You can ignore gravitational forces. (b) After what horizontal distance \(d\) does the proton return to its original elevation? (c) Sketch the trajectory of the proton. (d) Find the numerical values of \(h_{\max }\) and \(d\) if \(E=500 \mathrm{~N} / \mathrm{C}, v_{0}=4.00 \times 10^{5} \mathrm{~m} / \mathrm{s}\), and \(\alpha=30.0^{\circ}\)

In an experiment in space, one proton is held fixed and another proton is released from rest a distance of \(2.50 \mathrm{~mm}\) away. (a) What is the initial acceleration of the proton after it is released? (b) Sketch qualitative (no numbers!) acceleration-time and velocity-time graphs of the released proton's motion.

Point charges \(q_{1}=-5 \mathrm{nC}\) and \(q_{2}=+5 \mathrm{nC}\) are separated by \(3 \mathrm{~mm}\), forming an electric dipole. (a) Find the electric dipole moment (magnitude and direction). (b) The charges are in a uniform electric field whose direction makes an angle of \(37^{\circ}\) with the line connecting the charges. What is the magnitude of this field if the torque exerted on the dipole has magnitude \(7.5 \times 10^{-9} \mathrm{~N} \cdot \mathrm{m}\) ?

Two very large horizontal sheets are \(4.25 \mathrm{~cm}\) apart and carry equal but opposite uniform surface charge densities of magnitude \(\sigma .\) You want to use these sheets to hold stationary in the region between them an oil droplet of mass \(324 \mu \mathrm{g}\) that carries an excess of five electrons. Assuming that the drop is in vacuum, (a) which way should the electric field between the plates point, and (b) what should \(\sigma\) be?

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