# Why Does the Electric Field Component Ex Require a Negative Sign?

• yesmale4
In summary, the conversation discusses finding the correct answer for Ex and the use of minus in the solution. It is explained that the sign of the x-component depends on the sign of the linear charge density and the relative size of the bounds. The use of LaTeX in providing the solution is also encouraged.
yesmale4
Homework Statement
A uniformly charged rod is positioned along the x-axis as shown in the figure. Calculate the electric field at a point P on the y-axis.
Relevant Equations
dq=lamda*dx
de=kdq/r^2
hello i would like to understand to something.
here is the drew

now for my question:
i was able to find Ey and here is my correct answer:

when i try to find Ex i didnt understand something, i found the correct answer but i need to put minus before and i want to know why?
here is my solution for Ex (the correct answer):

yes i know i didnt solve the integral becuase first i want to know why there is minus

There is more charge to the right of the vertical line than to the left. The sum of all the contributions from the charge on the right will be grater than the contributions from the charge on the right. So if λ is positive the overall x-component will be negative and the opposite if λ is negative.

Of course, this should be built in the answer. You say that your answer is correct, but how do you know that? It looks like the ends of the rod are at ##x=-a## and ##x=\delta,## Your final answer should have no ##x## in it; only ##a## and ##\delta## and be the sum of two expressions resulting from the integral.

Also, please use LaTeX. Your handwriting is hard to decipher at times.

yesmale4 said:
when i try to find Ex i didnt understand something, i found the correct answer but i need to put minus before and i want to know why?
What is the sign of the x-component of the arrow you drew?

vanhees71 and Chestermiller
yesmale4 said:
why there is minus
Because the sign of the x component of the field due to an element dx at x has the sign of ##-\lambda x.dx##. I.e. if ##\lambda## is positive then where x is negative the field is to the right and vice versa.
kuruman said:
It doesn't really have an x in it - that's just a dummy variable. The OP has merely not done the final step of applying the bounds. Maybe that's what you meant.
And I think it is b, not ##\delta##.

haruspex said:
It doesn't really have an x in it - that's just a dummy variable. The OP has merely not done the final step of applying the bounds. Maybe that's what you meant.
Yes, that's what I meant. Applying the bounds would yield the difference of two terms in which case it would be easier to point out mathematically that the sign of the x-component depends on both the sign of the linear charge density and the relative size of the bounds.

## What is an electric field?

An electric field is a physical quantity that describes the influence of an electric charge on other charges in its vicinity. It is a vector field, meaning it has both magnitude and direction.

## How is the electric field of a charged rod calculated?

The electric field of a charged rod can be calculated using the equation E = kQ / L, where E is the electric field strength, k is the Coulomb's constant, Q is the charge on the rod, and L is the length of the rod.

## Does the electric field of a charged rod depend on distance?

Yes, the electric field of a charged rod decreases as the distance from the rod increases. This relationship follows the inverse square law, meaning that the electric field strength is inversely proportional to the square of the distance from the rod.

## What is the direction of the electric field of a charged rod?

The direction of the electric field of a charged rod is always perpendicular to the rod's length and points away from the positive end of the rod and towards the negative end. This direction can be determined using the right-hand rule.

## How does the electric field of a charged rod change if the charge or length of the rod is altered?

If the charge on the rod is increased, the electric field strength will also increase. Similarly, if the length of the rod is increased, the electric field strength will decrease. However, the relationship between these variables is not linear and is dependent on the inverse square law.

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