A particle with mass 2.30 g and charge +10.0 μC enters through a small hole in a metal plate wit

A particle with mass 2.30 g and charge +10.0 μC enters...

A particle with mass $2.30 \mathrm{g}$ and charge $+10.0 \mu \mathrm{C}$ enters through a small hole in a metal plate with a speed of $8.50 \mathrm{m} / \mathrm{s}$ at an angle of $55.0^{\circ} .$ The uniform $\overrightarrow{\mathbf{E}}$ field in the region above the plate has magnitude $6.50 \times 10^{3} \mathrm{N} / \mathrm{C}$ and is directed downward. The region above the metal plate is essentially a vacuum, so there is no air resistance. (a) Can you ignore the force of gravity when solving for the horizontal distance traveled by the particle? Why or why not? (b) How far will the $\Delta x,$ before it hits the metal plate?

Key Concepts

Electric Force

Electric force is the force experienced by a charged particle in the presence of an electric field, quantified by F = qE, where q is the particle's charge and E is the electric field. This force directly affects the particle's acceleration and trajectory and is essential in determining the motion of charged particles in uniform or non-uniform fields.

Uniform Electric Field

A uniform electric field has a constant magnitude and direction throughout a region. In problems involving charged particles, a uniform field simplifies the analysis since the electric force, and hence the acceleration due to the field, remains constant, allowing the use of standard kinematic equations to analyze motion.

Projectile Motion

Projectile motion describes the motion of an object moving under the influence of constant acceleration, typically gravity, and is characterized by independent horizontal and vertical components. When an additional constant force is present, such as from an electric field, the motion can still be analyzed using similar kinematic principles by considering the net acceleration in each direction.

Vector Decomposition

Vector decomposition involves breaking down an initial velocity into its horizontal and vertical components based on the given launch angle. This technique is critical because the horizontal motion remains unaffected by vertical forces (assuming no air resistance), while the vertical motion is influenced by acceleration due to forces like gravity and electric fields.

Relative Magnitude of Forces

Comparing the magnitudes of different forces, such as gravitational force and the electric force, is important to determine which force has a dominant effect on the motion. In scenarios where one force greatly exceeds another, the influence of the smaller force can sometimes be safely neglected to simplify analysis without significantly affecting accuracy.

Transcript

00:01 Hi, in the given problem here there is a horizontal plate which is having a hole in it like this.

00:12 An electric field in a direction vertically downward.

00:17 So there should be positive charge upward and negative charge downward as we know.

00:23 Electric field always goes away from positive and moves towards negative.

00:28 Then a charge, a positive charge comes out.

00:33 Here through this hole with a velocity v at an angle 55 degree so this velocity we can be resolved into two components here this is vx horizontal component and this vy vyi initial vertical velocity so first of all in the first part of the problem we have to discuss whether we should ignore the gravitational force acting on this charge particle or not.

01:11 Now the charge particle here is having a charge of 10 .0 microculum.

01:19 Its mass is 2 .30 gram.

01:23 It is moving with the speed v is equal to 8 .50 meter per second and electric field is 6 .50 into 10 -tish 3 .3.

01:35 Per 3 newton per column.

01:38 So first of all if we find its gravitational force means its weight, that is given by mg, means 2 .30 divided by 1000 into 9 .8 newton.

01:52 So this gravitational force will come out to be 22 .54 newton.

01:57 And then we find electrostatic force which is given by q into e.

02:02 Means this is 10 .0 into 20.

02:05 10 dash per minus 6 kulam into electric field 6 .50 into 10 dash par 3 newton.

02:12 This electrostatic force here will come out to be 65 and here it is 22 .54 into 10 dash per minus 3 newton and here also this is 65 .0 into 10 dash per minus 3 newton.

02:39 So as these two forces are comparable with each other...

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