An electron with a velocity of 10.0 m / s in the positive y-direction enters a region where ther

An electron with a velocity of 10.0 m / s in the positive...

An electron with a velocity of $10.0 \mathrm{m} / \mathrm{s}$ in the positive y-direction enters a region where there is a uniform electric field of $200 \mathrm{N} / \mathrm{C}$ in the positive $x$ -direction. What are the $x$ - and $y$ -components of the electron's displacement $2.40 \mu$ s after entering the electric-field region if no other forces act on it?

Key Concepts

Electric Force on a Charged Particle

This concept explains that a charged particle placed in an electric field experiences a force given by the relation F = qE, where q is the charge and E is the electric field. The direction of the force depends on the sign of the charge; for electrons, which have a negative charge, the force is opposite to the direction of the field.

Newton’s Second Law of Motion

Newton’s second law states that the net force on an object is equal to the mass of the object multiplied by its acceleration (F = ma). This principle is used to determine the acceleration of a charged particle when subjected to an electric force, as the acceleration is directly proportional to the force and inversely proportional to the mass.

Kinematics and Displacement

Kinematics involves the study of motion without considering its causes. For particles under constant acceleration, the displacement can be determined using equations such as s = v?t + 0.5at². This equation is applied separately to each component of motion when dealing with two-dimensional motion under a uniform field.

Component Analysis in Two-Dimensional Motion

In two-dimensional motion, the behavior of the object in each orthogonal direction (x and y) can be analyzed independently. This means one can calculate the displacement in the x-direction due to the acceleration from the electric field while the motion in the y-direction remains uniform because no net force is acting in that direction.

Transcript

00:02 Hi, in the given problem there is a uniform electric field along positive x -axis.

00:13 And a charge moving in a direction of positive y -axis enters into this electric field.

00:24 As you know, electric field is always from positive to negative.

00:28 So this charge particle which is actually an electron, then it will enter.

00:33 Into this electric field it will deflect towards the positivity means positive charge like this so we have to find the distance covered by this charge particle in the electric field means deflection of the chart particle in the electric field along y -axis and along x -axis here this deflection will be delta x along x y -axis delta y so the uniform electric field is given as 200 newton per column speed of electron along positive y -axis is given as let it be y and that is equal to 10 .0 meter per second now this electron will move a uniform motion along y -axis as it will experience the force and hence the acceleration only along x -axis.

02:22 The time given in which we have to find these deflections is p is equal to 2 .40 microsecond.

02:32 So in this one, much time that deflection covered the deflection undergone by this electron along y -axis is delta y is equal to distance equals to speed into time v y into delta t means this is 10 .0 meter per second multiplied by 2 .40 into 10 to minus 6 seconds so this delta y here comes out to be to 24 .0 or we can say this is 2 .40 this is it will become 2 .40 into 10 tisper minus 5 meter so this will be the distance the deflection undergone by the electron along y -axis now for its motion along x -axis there will be no initial component of the velocity along x -axis means v i x is equal to 0 and to find its acceleration along x -axis using the expression for the electrostatic force f equals to the product of charge with the electric field and also it is given as m a x where a x will be the acceleration along x -axis so this acceleration a .x will be equal to e e by m.

04:16 So as v i .x is equal to 0, a .x is equal to this much.

04:20 And time we know delta t is equal to 2 .40 microsecond...

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