Electric field in two dimensions

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SUMMARY

The discussion focuses on calculating the electric field at point P due to a positively charged rod using integration techniques. The electric field components are derived by breaking the rod into differential elements, represented as dL, and integrating the contributions from each segment. The key equations include Ey = (integrate)-L to +L Q/(4pi Epsilon(o) d^2)dq and dE = Ko*Lambda*dx/(r^2), where r is expressed as r = sqrt(L^2 + d^2). The conversation emphasizes the importance of considering both the x and y components of the electric field and applying trigonometric relationships to solve the problem.

PREREQUISITES
  • Understanding of electric fields and charge distributions
  • Familiarity with calculus, specifically integration techniques
  • Knowledge of Gauss's Law and its applications
  • Basic trigonometry for resolving vector components
NEXT STEPS
  • Study the application of Gauss's Law in different charge configurations
  • Learn about electric field calculations for continuous charge distributions
  • Explore integration techniques for vector fields in physics
  • Investigate the use of trigonometric identities in physics problems
USEFUL FOR

This discussion is beneficial for physics students, educators, and professionals involved in electromagnetism, particularly those working on problems related to electric fields and charge distributions.

stunner5000pt
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I have this problem which I am stuck on: Find an expression for the electric field at point P as in this figure
P
|
|
|
|
|
|
+++++++++++++++++++++++++++++++++++++++++++++++++++++++
<----------------------L-------------------------------->
Where the distance from P to the positive charged rod is Y.

All i know is that you need to break the positive charged rod into little differentials such as dL and integrate something like this -
Ey = (integrate)-L to +L Q/(4pi Epsilon(o) d^2)dq
However i am not sure about the X components and how taht is calculated, please help.
 
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Why don't u apply Gauss Th
 
Well this one can be tricky. Now with your picture there and then your claim I am confused, is the point located perpendicular to the end of the line of charge?
If that is the case then integrating along the length from 0 to L might be easier...
anyway
if you look at the picture, as you move from the left to the right, and angle is formed with the line extending from P along L. as you move down L this angle changes.
As the angle becomes larger the radius changes as well.
dEx=dEcos(theta)
dEy=dEsin(theta)
Now dE=Ko*dq/(r^2)
substituting the charge density lambda*dx for dq we get
dE=Ko*Lambda*dx/(r^2)
if you replace r with the varibles we do know you get
r=sqrt(L^2+d^2)
where d is the vertical distance from the line to the point.
now making the necessary trigonometric changes you can find the the integral for each component.

Hope that helps
 
The book claims the answer is that all the magnitudes are the same because "the gravitational force on the penguin is the same". I'm having trouble understanding this. I thought the buoyant force was equal to the weight of the fluid displaced. Weight depends on mass which depends on density. Therefore, due to the differing densities the buoyant force will be different in each case? Is this incorrect?

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