Electric field of a fish An African fish called the aba has a charge q=+1.0 ×10^-7 C at its head

Electric field of a fish An African fish called the aba has...

Key Concepts

Coulomb's Law

Coulomb's law is the fundamental principle that quantifies the force or field between two point charges. It states that the electric force is directly proportional to the product of the two charges and inversely proportional to the square of the distance between them. In any medium, the effect of the environment is accounted for by the medium's permittivity, which alters the strength of the interaction.

Electric Dipole

An electric dipole is composed of two equal but opposite charges separated by a finite distance. This arrangement creates an electric field that, especially in regions far from the dipole, decreases with the cube of the distance from the dipole. The concept of a dipole is essential for understanding the spatial variation of the field produced by such charge configurations.

Electric Field

The electric field is a vector field that represents the force experienced per unit charge at any given point in space. It is produced by charges and defined in terms of both magnitude and direction, providing a way to understand how a test charge would interact with the surrounding charge distribution.

Force on a Charge

The force acting on any charge in an electric field is determined by the simple relation F = qE, where q is the charge and E is the electric field at the location of the charge. This direct relationship allows for the calculation of the force exerted on a charge by any known electric field.

Dielectric Medium

A dielectric medium, such as water, affects the electric interactions between charges by reducing the strength of the electric field compared to what it would be in a vacuum. This reduction is quantified by the medium's dielectric constant, which modifies Coulomb's law by effectively increasing the distance between interacting charges, thereby diminishing the force and field magnitude.

Transcript

00:01 In this question, we're going to find the electric field at a certain point by an african fish called an abba that creates an electric field.

00:11 And then we're going to find the force on an ion at that point.

00:18 And note that the fish and the ion are in water.

00:23 So that adds, oops, there we go.

00:26 Okay.

00:27 So this is going to add a little something to our equations.

00:36 We need to consider the dielectric constant of this material.

00:41 And what's going to happen is instead of just calculating electric field as kq over r squared, we're going to need to add that kappa dielectric constant in, um, that it goes into the denominator.

01:05 A kappa for water given on page, i think it was 555 of the textbook.

01:10 Don't quote me on that, uh, was 81.

01:13 So let's look at this.

01:25 Um, so remember that electric fields are defined by what a positive test charge would feel at a particular point.

01:34 So our electric field contribution from the negative charge is going to be down to the left and our electric field contribution from the positive charge at the fish's head is going to be straight up.

01:51 And notice that since, uh, the positive charge is closer, it is going to contribute more to this total electric field.

02:01 And one more thing that we're going to need in all of this process is the distance from, um, getting straight line very well, but we're going to need the distance from, uh, the negative charge at the fish's tail to, um, i'm going to call it r minus to that point.

02:26 And we can use the pythagorean theorem for this.

02:29 So r minus will be equal to the square root of 0 .12 squared plus 0 .16 squared, which gives me 0 .2 meters.

02:52 All right.

02:57 Um, so now that we have knocked that out, um, now we can go ahead and dive into this actual calculation.

03:10 So because this is a two -dimensional situation, we need to find the electric field x component separately from the y component.

03:21 And that x component is going to come exclusively from the x component of the field from the negative charge.

03:33 So i'm going to call it e minus x.

03:36 And we're going to calculate that with kq over kappa r minus squared times the cosine of the angle from the horizontal, which i haven't dashed in yet.

03:58 So let's go dash that in.

04:02 So it's going to be this angle here, which also happens to be, if we were to construct a right triangle within the fish, it's going to be the same as the angle that r minus makes with the line that cuts through the body of the fish.

04:22 That's kind of nice because that lets me specify that the cosine of this angle is simply going to be, um, the, uh, adjacent side 0 .16 over the hypotenuse, which was r minus, which is 0 .2...

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