Thin Charged Isolated Rod -- Find the electric field at this point

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The discussion focuses on calculating the electric field generated by a thin charged isolated rod at a specific point. The user acknowledges the need for integration due to the rod's continuous charge distribution, unlike point charges or spheres. Key steps involve defining the charge element, labeling positions, and setting up the coordinate system correctly. Participants emphasize the importance of using symbols rather than numbers initially and suggest deriving the electric field components through integration. The conversation highlights the need for clarity in the setup to facilitate accurate calculations.
Ugnius
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Homework Statement
Thin isolated rod , 1 meter in length , placed on Y axis , find rod's created electic field to point P(1,0.75) , rod's linear charge density lambda = 6.8 micro coulumbs per meter.
Relevant Equations
dQ=Q/Ldx
Hi , I've been trying to manage a solution in my head and i think I'm on the right path , i just need some approval and maybe some tips.
So it's obvious I can't solve this without integration because law's only apply to point charges , and i can't shrink this object to a point as i could do with sphere. I've looked everywhere and couldn't find a formula that would involve anything like this atleast in my native language. If the point was in the middle of the rod and all the other vectors would cancel out , i could just calculate Ex = E , but now it's offset and I'm confused
 

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Put the origin of coordinates at the midpoint of the rod. Consider an element ##dq= \lambda~dy'## on the rod at location ##y'## from the origin. Can you find the x and y components of the electric field due to this element alone at point P? Call them ##dE_x## and ##dE_y##. To find the net field, you need to do two separate integrals over ##y'##.

Start writing things down, post them and we will help you through. Please use symbols not numbers until the very end. Call the coordinates of P ##x_P## and ##y_P##.
 
kuruman said:
Put the origin of coordinates at the midpoint of the rod. Consider an element ##dq= \lambda~dy'## on the rod at location ##y'## from the origin. Can you find the x and y components of the electric field due to this element alone at point P? Call them ##dE_x## and ##dE_y##. To find the net field, you need to do two separate integrals over ##y'##.

Start writing things down, post them and we will help you through. Please use symbols not numbers until the very end. Call the coordinates of P ##x_P## and ##y_P##.
1632610131650.png

I don't know if I managed to understand what you meant , but is it something like this?
 
Yes, something like this. To avoid confusion:
1. Use y or y' for positions along the y-axis.
2. Define distances from the origin which you chose to be at the end of the rod. That's fine.
3. The length element should be labeled ##dy## and its position should be labeled ##y## from the end of the rod (point O).
4. Point C should be at {0, ##\frac{3}{4}L##} and point P should be at {{x, ##\frac{3}{4}L##}.
Got it? Please redraw the figure according to the above so that we can refer to it in the future. Thanks.

Now you need to find an expression first for the magnitude of ##dE## and then worry about the components. Use the expression ##dE= \dfrac{dQ}{4\pi\epsilon_0r^2}## for a point charge.

Look at the new drawing. Your point charge is the length element ##dy##. What is the charge ##dQ## on it? What is the magnitude of the field generated by it at point P in terms of the symbols in the drawing?
 
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My attempt: Initial total M.E = PE of hanging part + PE of part of chain in the tube. I've considered the table as to be at zero of PE. PE of hanging part = ##\frac{1}{2} \frac{m}{l}gh^{2}##. PE of part in the tube = ##\frac{m}{l}(l - h)gh##. Final ME = ##\frac{1}{2}\frac{m}{l}gh^{2}## + ##\frac{1}{2}\frac{m}{l}hv^{2}##. Since Initial ME = Final ME. Therefore, ##\frac{1}{2}\frac{m}{l}hv^{2}## = ##\frac{m}{l}(l-h)gh##. Solving this gives: ## v = \sqrt{2g(l-h)}##. But the answer in the book...

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