Force on a charge from a nearby charged rod

AI Thread Summary
The discussion focuses on calculating the force on a charge due to a nearby charged rod using integration. The user initially presents their approach, integrating the force equation but expresses confusion regarding the appearance of R^(-3) in a mentor's solution. The relevant equations involve the electric field's magnitude and direction, which are crucial for correctly setting up the integral. Clarification is sought on how to properly account for vector addition in the integration process. Understanding these concepts is essential for accurately solving the problem.
cherry
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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.
Screenshot 2024-01-28 at 3.25.30 PM.png


Here is where I start getting confused.
I looked at a different physics forum post and the mentor gave this equation:
Screenshot 2024-01-28 at 3.27.56 PM.png


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!
 
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cherry said:
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.
 
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?
 
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