# Why do the Horizontal components cancel in E-Field

• iceman_ch
In summary, in the given scenario, a uniform charge density exists on Plane X=0 while on Plane X=a, the uniform charge density is negative. The task is to determine the electric field intensity in the region X<0. The solution involves integrating the different components of the equation and converting to polar coordinates to simplify the process. The final equation shows that the electric field in the region X<0 is equal to 0.
iceman_ch

## Homework Statement

Plane X=0 has a uniform charge density, $$\rho$$s while plane X=a has -$$\rho$$s. Determine the electric field intensity in region X<0.

## Homework Equations

E=$$\frac{Q}{4\Pi\epsilon0R2}$$aR

## The Attempt at a Solution

I already know the answer to this question. What I'm trying to do is figure out a small part of the math behind it. Let's start by looking at one plate instead of two. How can I prove mathematically that all of the horizontal components cancel out leaving behind the component pointing away from the plate. I understand why but I need to understand the math behind it. This makes working problems so much easy when I can understand the math behind it easily. So here goes what I tried to do.

First I need to define a point to examine my e-field at. r=(x,y,z) next i need to define the different parts of the equation above. Best way to do this is to look at very very small peices and then adding them up.

dE = $$\frac{dQ}{4\Pi\epsilon0R2}$$
dQ = $$\rho$$sdS
dS = dY dZ
R = ?

I want to put that R=r-r' and that r' would the location of the very very small plate which would be (0,y',z') which would make R=(x,(y-y'),(z-z')) The y and z would stay constant because they are the point that we are evaluating E at. This is were I get lost. I know technically that y and z can be ignored because they will add to zero but, how can I prove this. If I keep going with my attempt the next step would be to find the magnitude of R and the unit vectors of R. I'm going to leave these out for now since they are long to type. So now I would put everything together and integrate. This is where everything gets messy. I know that if I put in a value for y say 4 and then integrate (4-y') from 0 to 8 the answer is 0. How can I write this out when even though y is constant because I don't know what it is I can't really prove that it ends up being 0.

I worked on the problem a little bit more and I believe that I've worked out the correct solution. First when you write everything out it takes up a lot of paper and it is somewhat complicated. So the first thing is to go ahead and integrate the y and z components since they are the horizontal components that should go away by the end of the solution. If you do this they do indeed go to zero. I'll write out the y component so that it's easier to understand what I'm writing.

ay=$$\frac{\rhos}{r\Pi\epsilon0}$$$$\int\int\frac{(y-y')}{(x2+(y-y')2+(z-z')2)\frac{3}{2}}}$$

This ends up equaling zero because the integral from infinity to negative infinity of (y') is 0. So this leaves the X component as the only one that has to be deal with. The X component is

ax=$$\frac{\rhos}{r\Pi\epsilon0}$$$$\int\int\frac{(x)}{(x2+(y-y')2+(z-z')2)\frac{3}{2}}}$$

First thing to do to evaluate this integral is to realize that X is constant and can be relabeled H to be a little less confusing. The next thing if you draw everything out you realize that converting to polar coordinates would be a whole heck of a lot easier. This changes the equation to look like

ax=$$\frac{\rhosH}{r\Pi\epsilon0}$$$$\int\int\frac{(1)}{(H2+\rho2)\frac{3}{2}}}$$$$\rho$$ d$$\rho$$ d$$\Phi$$

The limits of integration would be 0 to infinity and 0 to 2pi.

Integrating this will leave

$$\frac{2\pi}{H}$$

Which makes the final equation of

$$\frac{\rhos}{2\epsilon0}$$

Then going back to the original problem and noticing that H isn't involved in the final equation means that the e-field in the region X<0 is equal to 0. My textbook does point the final equation out but, I don't like using these equation without understanding how they were derived. Hopefully I did all of the work correctly.

## Why do the Horizontal components cancel in E-Field?

The cancellation of horizontal components in an electric field is a fundamental concept in electromagnetism. It is often misunderstood and raises questions such as:

## 1. What is the concept of horizontal component cancellation in an electric field?

In simple terms, the cancellation of horizontal components in an electric field means that the total electric field at a point is only determined by the vertical components of all the individual electric fields acting on that point.

## 2. Why do horizontal components cancel in an electric field?

The cancellation occurs due to the vector nature of electric fields. Electric fields have both magnitude and direction, and when two electric fields with equal magnitude but opposite direction (horizontal components) act on a point, they cancel each other out, leaving only the vertical component.

## 3. Is the cancellation of horizontal components the same as the superposition principle?

No, the superposition principle states that the total electric field at a point is the vector sum of all the individual electric fields acting on that point. The cancellation of horizontal components is just one aspect of the superposition principle.

## 4. Does the cancellation of horizontal components only occur in certain situations?

No, the cancellation of horizontal components occurs in all situations where there are two or more electric fields acting on a point. This can be seen in simple examples such as parallel plate capacitors or more complex situations involving multiple charges.

## 5. How does the cancellation of horizontal components affect the strength of the electric field?

The cancellation of horizontal components does not affect the strength of the electric field at a point, as the magnitude of the total electric field is determined by the vertical components of all the individual electric fields acting on that point. The cancellation only affects the direction of the electric field at that point.

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