# Calculate the electric field due to a line of charge of finite length

• Yalanhar
In summary, the problem involves finding the electric field at point P, where a thin rod of length L and charge Q is uniformly charged with linear charge density ##\lambda = q/l##. Using the equation ##dE=\frac{Kdq}{r^2}## and solving for the electric field in the x and y direction, Tipler's solution involves using negative signs for the x component and the overall expression for the y component due to the chosen coordinate system. This is because the quantity ##x## is negative and can be written as ##|x| = -x## to obtain a positive value.
Yalanhar
Homework Statement
A thin rod of length L and charge Q is uniformly charged, so it has a linear charge
density ##\lambda =q/l## Find the electric field at point where is an arbitrarily positioned
point.
Relevant Equations
##dE=\frac{Kdq}{r^2}##
Homework Statement: A thin rod of length L and charge Q is uniformly charged, so it has a linear charge
density ##\lambda =q/l## Find the electric field at point where is an arbitrarily positioned
point.
Homework Equations: ##dE=\frac{Kdq}{r^2}##

A thin rod of length L and charge Q is uniformly charged, so it has a linear charge
density ##\lambda =q/l## Find the electric field at point P where P is an arbitrarily positioned
point.

Tipler solves this one. But there are 2 things that I do not understand about his solution.

1) Why on 5 he used the negative sign for Xs? shouldn't he use Xs positive as in:
##tan\theta =\frac {y}{x}##
##x = \frac{y}{tan\theta}##
##dx = -ycsc^2\theta##

But by doing this my Ex gets negative.
2) What the negative sign on Ey means? The orientation is opposite to the unit vector j?

Sorry for bad image and english.

Yes, dealing with the signs in this calculation can be annoying.

1. Note that ##\theta## is between 0 and ##\pi/2##. So, ##\tan \theta## is positive. But ##\frac {y_p}{x_s}## is negative (why?). So, it would not be correct to write ##\tan \theta = \frac{y_p}{x_s}##.

2. In the final expression for ##E_y##, there is an overall negative sign. However, you also need to consider the sign of the expression inside the parentheses.

Yalanhar
(1) Yes, but x is positive or negative? Tipler used negative so the derivative can be positive. But shouldn't i use modulus?

(2) I see it now. ##cos\theta_{2} - cos\theta_{1}## is negative. So double negative gives me positive.

Yalanhar said:
(1) Yes, but x is positive or negative? Tipler used negative so the derivative can be positive. But shouldn't i use modulus?
I'm not sure I follow you here.

Do you agree with Tipler's expression ##\tan \theta = \frac{y_p}{|x_s|}##?

If so, then note that Tipler's choice of the coordinate system implies that ##x_s## is a negative number. So, ##|x_s| = -x_s##.

TSny said:
I'm not sure I follow you here.

Do you agree with Tipler's expression ##\tan \theta = \frac{y_p}{|x_s|}##?

If so, then note that Tipler's choice of the coordinate system implies that ##x_s## is a negative number. So, ##|x_s| = -x_s##.
For example,

in this case ##\theta = arctan(\frac {0.5}{0.87})## and not ##\theta = arctan(\frac {0.5}{(-0.87)})##

##\tan \theta = \frac{y_p}{|x_s|}##?I agree

##|x_s| = -x_s##. is this true because it's implied that x is negative, then he used -x to be positive? It makes sense

Yes. Whenever you have a quantity ##x## that is negative, then you can write ##|x| = -x##. Both sides of this equation are positive numbers.

Yalanhar
TSny said:
Yes. Whenever you have a quantity ##x## that is negative, then you can write ##|x| = -x##. Both sides of this equation are positive numbers.
Thanks ! i finally get it

## What is the formula for calculating the electric field due to a line of charge of finite length?

The formula for calculating the electric field due to a line of charge of finite length is E = (λ/4πε0)(1/r1 - 1/r2), where λ is the linear charge density, ε0 is the permittivity of free space, r1 is the distance from the beginning of the line of charge, and r2 is the distance from the end of the line of charge.

## What is the linear charge density and how is it related to the charge of the line?

The linear charge density (λ) is the amount of charge per unit length along the line of charge. It is related to the total charge (Q) of the line by the equation λ = Q/L, where L is the length of the line of charge.

## How does the electric field vary along the line of charge?

The electric field varies inversely with the distance from the line of charge. This means that the electric field is strongest closest to the line and decreases as the distance from the line increases.

## Can the electric field due to a line of charge of finite length be zero?

Yes, the electric field can be zero at certain points along the line of charge. This occurs when the distance from the line of charge is equal to the length of the line, or when r1 = r2.

## What is the direction of the electric field due to a line of charge of finite length?

The direction of the electric field is always perpendicular to the line of charge. This means that the electric field points away from the line for positive charges and towards the line for negative charges.

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