A small object with mass m, charge q v0=5.00 ×10^3 m / s is projected into a uniform electric f

A small object with mass m, charge q v0=5.00 ×10^3 m / s is...

A small object with mass $m$, charge $q$ $v_{0}=5.00 \times 10^{3} \mathrm{~m} / \mathrm{s}$ is projected into a uniform electric field between two parallel metal plates of length $26.0 \mathrm{~cm}$ (Fig. $\mathbf{P} 21.78$ ). The electric field between the plates is directed downward and has magnitude $E=800 \mathrm{~N} / \mathrm{C}$. 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 $d=1.25 \mathrm{~cm}$ from its original direction of motion and reaches a collecting plate that is $56.0 \mathrm{~cm}$ from the edge of the parallel plates. Ignore gravity and air resistance. Calculate the object's charge-to-mass ratio, $q / \mathrm{m}$.

Key Concepts

Charge-to-Mass Ratio Calculation

By relating the measured deflection of a particle in the vertical direction to its horizontal travel time and the known electric field strength, the charge-to-mass ratio (q/m) can be determined. This involves using the kinematic equation for motion with constant acceleration to solve for the acceleration (a = qE/m) and then rearranging terms to isolate q/m based on the observed displacement and the time the particle spends under acceleration.

Projectile Motion and Kinematics

The analysis of the particle’s motion involves separating it into horizontal and vertical components. The horizontal component of motion is unaffected by the electric field and follows constant velocity motion, while the vertical component experiences constant acceleration due to the electric force. These independent kinematic principles allow one to predict displacement and time of flight in each direction using standard kinematic equations.

Uniform Electric Field

In a uniform electric field, the magnitude and direction of the electric force are constant. A charged particle placed in such a field experiences a constant force equal to the product of its charge and the field strength (F = qE). This concept is fundamental when considering how an electric field influences the motion of charged particles, as the force results in a constant acceleration according to Newton’s second law.

Electric Force on a Charged Particle

The force exerted on a charged particle by an electric field is described by the Lorentz force equation, where for a particle moving in an electric field (with no magnetic field present) the force is F = qE. This explains how the particle is accelerated by the field, and how its mass and charge determine the magnitude of this acceleration (a = qE/m).

Transcript

00:01 Here at the problem we have a particle of charge q and mass m that is moving the possible x -ivection with initial velocity v0 which is equal to the 5 times 10 per 3 meter per second i had through a parallel plate capacitor of length 26 centimeters then it moved it moved long up to 56 centimeters up to a screen and there was a deflection from the horizontal central axis with the distance d which is equal to 1 .25 centimeters and through the parallel plate capacitor there was an electric field e which is pointing downwards of magnitude 800 volts per meter and we need here in this problem to calculate the ratio between the charge and the mass of this particle so here here we start by getting the time at t1 that that was spanned by this object in the parallel plate capacitor so here t1 which represents that time is equal to the distance over the velocity so here the distance is just 26 times 10 power negative 2 meters divided by the velocity which is 5 times 10 power 3 and this is equal to 52 times 10 power negative 6 seconds and here according to kulom's law we have the force just equal to the charge multiplied by the electric field and this is equal to an a according to newton's law so here the acceleration of the particle just equal to q over m multiplied by the electric field or 800 multiplied by q over m and i will represent q over m by a constant k so this is equal to 800k and the acceleration here is pointing downwards so we can apply new to the low dynamics so here the vertical distance that this particle has traced and the parallel plate capacitor will be represented by s1 so here according to newton's law of motion the initial velocity in the vertical direction is zero so here s1 just equal to one half times the acceleration which is 800k times t1 squared which is 52 times 10 bar negative 6 all squared and this is equal to 1 .352 times 10 0 .0.

03:43 K this is for s1 and now we get the time t2 which is the time spanned by this object outside of the parallel plate capacitor so here the horizontal distance is just 56 centimeters divided by the velocity which is 5 times 10 power 3 and this would be equal to 112 times 10 or negative 6 seconds and here in this problem we ignore the gravity so now the vertical motion outside the parallel plate capacitor is just equal to s2 which would represent this distance or this vertical distance and this is equal to some velocity v times t2 and here v is the final vertical velocity with which the particle came out of the parallel plate capacitor so here or but here v is just equal to the the initial the last in the way direction is just 0 so this is just equal to a t1 so now s2 is just equal to a t1 multiplied by t2 and this is equal to 800 k multiplied by t1 multiplied by t1 which is 52 times 10 power negative 6 times t2 which is 112 times 10 bar negative 6 and s2 is just equal to 4 .6592 times 10 bar negative 6 k.

06:06 This is the vertical distance traced by the particle outside of the parallel plate capacitor.

06:13 So now the deflection distance d is just equal to s1 plus s2 and this is equal to 4 .6592 times 10 .5 .6 plus 1 .352 times 10 .5 .2 times 10 .5 .9...

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