How Does Electric Flux Change with Dipole Orientation and Distance?

In summary, the problem involves a small electric dipole located a distance z from the center of a loop with its dipole moment vector directed along the loop's axis. The electric field component perpendicular to the loop at any point P is given by the formula E(perpendicular) = p/(4 pi (epsilon) r^3) (3cos^2(theta) - 1), where (theta) is the angle between the loop's axis and the line from the dipole to P. The electric flux through the loop is then calculated using the formula flux = integral E(perpendicular) dA and is equal to [p/(2(epsilon))] a^2/(z^2 + a^2)^3/2. The
  • #1
Eric_meyers
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Homework Statement


A very small electric dipole is a distance z from the center of a loop of radius a, with its dipole moment vector p directed along the axis of the loop. The component of the electric field of the dipole perpendicular to the loop at any point P is a distance r from the dipole given by

E(perpendicular) = p/(4 pi (epsilon) r^3) (3cos^2(theta) - 1)

-Where (theta) is the angle between the axis of the loop and the line from the dipole to P. Show that the electric flux through the loop is given by.

flux = [p/(2(epsilon))] a^2/(z^2 + a^2)^3/2


Homework Equations


flux = integral E(perpendicular) dA


The Attempt at a Solution



Ok, I've been at it for an hour and a half, and really from what I can surmise is this.

The loop is a circle whose center is at the origin, the dipole above it - a distance z, has a point a distance r from it. The cos factor in the electric field given to me corrects the field propagating from the dipole to only include the perpendicular components to my loop.

cos(theta) = z/r where r can't = 0

my area of my loop = pi * a^2

I'm having trouble thinking if this is an integral problem or if I can apply Guass's law and simply just multiple my electric field by my area to get the flux.

if I multiply the area with the corrected cos factor I get. 3cos^2(theta) - 1 = 3* z^2/r^2 - 1

3 * z^2 * p * pi * a^2/ (4 * pi * epsilon * r^3) [[[a being the radius of loop]]]

Which would then simplify to

3/4 * z^2 * p * a^2/(epsilon * r) and r would be (a^2 + z^2)^1/2

But clearly this is not right.
 
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  • #2
So I assume it's an integral.I'm having trouble understanding how to set up the integral, so any help would be greatly appreciated. I assume my limits of integration, right? r= 0 to a theta = 0 to 2piI'm just not sure how to write this.
 
  • #3




Dear student,

Thank you for your question. It seems like you have made some progress in understanding the problem, but you are still unsure about how to approach it. Let me try to help you.

First, let's review the problem statement. We have a small electric dipole located at a distance z from the center of a circular loop with radius a. The dipole moment vector is directed along the axis of the loop. We are interested in finding the electric flux through the loop, which is given by the integral of the perpendicular component of the electric field over the loop's area.

Now, you correctly identified the expression for the perpendicular component of the electric field at a point P located at a distance r from the dipole. However, you are missing the integral part of the problem. In order to find the flux through the loop, we need to integrate the electric field over the loop's area. This is given by the surface integral:

flux = ∫∫ E(perpendicular) dA

where dA is an infinitesimal element of the loop's area.

So, how do we perform this integral? One way is to use Gauss's law, which states that the electric flux through a closed surface is equal to the total charge enclosed by that surface divided by the permittivity of free space (ε0). In our case, the loop is a closed surface, and the total charge enclosed is equal to the dipole moment p. Therefore, we can write:

flux = p/ε0

But this is not the same expression that we were asked to show. So, let's see if we can derive the given expression from this result.

The key is to relate the distance r from the dipole to the distance z from the center of the loop. We can do this by using the Pythagorean theorem, which tells us that r^2 = z^2 + a^2. Substituting this into our expression for the perpendicular component of the electric field, we get:

E(perpendicular) = p/(4πεr^3) (3cos^2θ - 1)

= p/(4πε(z^2 + a^2)^3/2) (3(z^2/(z^2 + a^2)) - 1)

= p/(4πε(z^2 + a^2)^3/2) (2z^2/(z^2
 

Related to How Does Electric Flux Change with Dipole Orientation and Distance?

1. What is electric flux?

Electric flux is a measure of the amount of electric field passing through a given surface. It is defined as the product of the electric field and the area of the surface, and is represented by the symbol Φ.

2. How is electric flux calculated?

The electric flux through a surface can be calculated by taking the dot product of the electric field and the surface area vector. The formula is Φ = E⋅A⋅cosθ, where E is the electric field, A is the surface area, and θ is the angle between the electric field and the surface area vector.

3. What is a “tough” electric flux problem?

A “tough” electric flux problem is a problem that requires the use of advanced mathematical and analytical skills to solve. This can include problems with complex geometries, varying electric fields, or non-uniform surface areas.

4. What are some common strategies for solving tough electric flux problems?

Some common strategies for solving tough electric flux problems include using vector calculus, Gauss’s law, and symmetry arguments. It is also helpful to break the problem down into smaller, more manageable parts and to draw diagrams to visualize the problem.

5. How can understanding electric flux be useful in the real world?

Understanding electric flux can be useful in a variety of real-world applications, such as designing electrical circuits, analyzing the behavior of electric fields in different materials, and studying the effects of electric fields on the human body. It is also an important concept in the study of electromagnetism and can help scientists and engineers solve complex problems in these fields.

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