A small sphere with charge q=5.00 μC and mass 0.500 g is traveling horizontally toward the east

A small sphere with charge q=5.00 μC and mass 0.500 g is...

A small sphere with charge $q=5.00 \mu \mathrm{C}$ and mass $0.500 \mathrm{~g}$ is traveling horizontally toward the east at a height of $60.0 \mathrm{~cm}$ above the ground. The sphere has a speed of $2.00 \mathrm{~m} / \mathrm{s}$ as it enters a region of uniform electric field with magnitude $E$. What is $E$ if the sphere has a speed of $5.00 \mathrm{~m} / \mathrm{s}$ just before it strikes the ground?

Key Concepts

Projectile Motion

Projectile motion describes the trajectory of an object that moves under the influence of gravity while having an initial horizontal velocity. In problems where additional forces (like an electric force) act on the particle, the motion can be analyzed by separating the horizontal and vertical components, allowing each to be considered independently.

Gravitational Potential Energy

Gravitational potential energy represents the energy stored by an object due to its position in a gravitational field. When an object falls, this potential energy is converted into kinetic energy, and it must be accounted for when analyzing energy transformations in systems that include both gravitational and other forces such as electric forces.

Work-Energy Principle

This principle connects the net work done on an object to its change in kinetic energy. In problems involving forces like gravity and electric fields, the work-energy principle is used to equate the total work done by these forces over a displacement to the difference between the final and initial kinetic energies of the object.

Electric Force

An electric field exerts a force on any charged particle defined by F = qE. This force contributes work to a particle's mechanical energy when the particle moves along a path that has a component in the direction of the force, thereby altering the particle’s speed and kinetic energy.

Transcript

00:01 Hello, everyone.

00:01 In this problem, we're asked to find the electric field magnitude for the follow scenario.

00:07 We have a five times into minus six quorum charge or five microculeum charge of mass 0 .500 grams, so 5 times into minus 4 kilograms, moving at a height of 0 .60 meters with an initial velocity of 2 meters per seconds east, and it's entering a uniform electric field of some sort and has a final speed of 5 meters per seconds as it hits the ground.

00:37 So we know that the acceleration due gravity is 9 .8 meters per second squared and we're looking for once again the electric field magnitude.

00:47 So actually there is an assumption that you have to make here because just telling you that the final speed is 5 meters, meters per seconds, there are many ways that you could engineer the electric field depending on its direction and its magnitude such that that would be true.

01:04 So, you know, this five meters per second square is the combination, is the sum of the x and y components of the final velocity.

01:12 And so those components can be pretty much arbitrary unless you make an assumption.

01:18 So here i'm going to assume that the electric field is horizontal and then we'll see where that gets us.

01:24 So with the assumption, we can start by finding the time it takes for the charge to hit the ground.

01:32 And that, of course, is unaffected by the electric field.

01:35 This is pointing in the horizontal direction.

01:39 So it's not providing any force in the vertical direction for the charge.

01:45 So we know that the initial y speed or y velocity is near zero meters per second squared or zero meters per second.

01:54 So we can just work out using the height what the time it takes for it as charged to fall to the ground is.

02:01 So we're just going to use our basic kinematic equation of the height or y is equal to, you know, initial y position plus, or, you know, final y position is equal to initial y position plus initial y velocity or right speed times time plus a half times acceleration in the y direction times d squared.

02:20 Or, of course, the first two terms are zero.

02:22 And so we just know that it falls through a height of h under purely the action of the acceleration due to gravity.

02:30 So the acceleration is g.

02:31 And we just have that t we're except to be the square root of 2 times h over g, which if you calculate it, that's 0 .35 seconds...

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