Force on a charge from a nearby charged rod

[cherry]

Homework Statement

An electric charge of Q = 6.4 C is distributed uniformly along a rod of length 2L, extending from y = -15.6 cm to y = +15.6 cm, as shown in the diagram. A charge q = 3.05 C, and the same sign as Q, is placed at (D,0), where D = 44 cm. Integrate to compute the total force on q in the x-direction.

Relevant Equations

dF=(kqQ/2Lr^2)dy

Hi! My attempt at this solution was:

∫dF = k*q*Q / 2L ∫ (1/r^2) dy

and we know that r^2 = D^2 + y^2 based on the diagram.

Here is where I start getting confused.
I looked at a different physics forum post and the mentor gave this equation:

I am mainly confused with the math.
How did he end up with R^(-3) inside the integral?
Otherwise, I ended up solving the integral and getting the right answer.

Thanks!

 

[SammyS]

Homework Statement: An electric charge of Q = 6.4 C is distributed uniformly along a rod of length 2L, extending from y = -15.6 cm to y = +15.6 cm, as shown in the diagram. A charge q = 3.05 C, and the same sign as Q, is placed at (D,0), where D = 44 cm. Integrate to compute the total force on q in the x-direction.
Relevant Equations: dF=(kqQ/2Lr^2)dy

Hi! My attempt at this solution was:

∫dF = k*q*Q / 2L ∫ (1/r^2) dy

and we know that r^2 = D^2 + y^2 based on the diagram.
View attachment 339345

Here is where I start getting confused.
I looked at a different physics forum post and the mentor gave this equation:
View attachment 339346

I am mainly confused with the math.
How did he end up with R^(-3) inside the integral?
Otherwise, I ended up solving the integral and getting the right answer.

Thanks!

Look at Post #3 in the thread that you gave the link for.

 

[kuruman]

The electric field at point D due to charge ##dq## is a vector that has a magnitude and a direction.
As shown in the figure, the magnitude is ##dE=\dfrac{kQ~dy}{2L(D^2+y^2)}##.
In what direction does this vector contribution point?
An integral is an addition. How do you add vectors?