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lie segment with linear charge density
lie segment with linea  
I am havig some difficult regarding this problem. I do not understand what need to be done. Any help would be greatly appreiated. The image displays what is being reprsented.

Consider a finite line segment having a linear charge density lamda(x) = 10 |x| (x is in m and lamda is in C/m). The line segment extends to a length of L (not shown) on either side of the origin.

Hint: In setting up the integrals, break the rod into two symmetric parts.

(a) Consider a small element of length delta x and position x. What is the charge delta q of this element? Express your answer in terms of lamda, x and delta x.
(b) What is the total charge of the rod? Express your answer in terms of lamda and L.
(c) What are (i) the electric field delta E, (ii) the electric potential delta V, at P due to delta q? Express your answers in terms of lamda, x, delta x and y.
(d) What are (i) the net electric field E, (ii) the net electric potential V, at P due to the entire rod? Use symmetry wherever possible, and evaluate your final integrals. Express your answers in terms of lamda, y and L.

(A) The charge of the element of charge dQ will be equal to the charge per unit length, lambda, multiplied by the length of the little piece of length dx: dQ=lambda*dx.
(B) The total charge of the entire rod will be equal to the charge per unit length, lambda, multiplied by the entire length L of the rod: q=lambda*L.
(C) The electric field dE at point P due to the little charge element dQ can be described by Coulomb's Law: dE=k*dQ/r^2=k*dQ/(x^2+y^2)
While the electrostatic potential contribution dV at point P will be: dV=k*dQ/Sqrt(x^2+y^2)
(E) The total electric field at point P will be pointing in the positive y direction since the charge distribution is positive and since for every charge element dQ on the left side there is an additional charge element dQ on the right side whose x components are equal but opposite and therefore cancel out. Meanwhile the y components both point in the positive y direction and, therefore, add. Adding all of the electric field contributions in the positive y direction calls for integrating the electric field contributions of all of the little pieces of charge dQ in the y direction:
Ey=Integral(k*dQ*y/(x^2+y^2)^3/2) evaluated between -L/2 and +L/2
This integral becomes:
Ey=Integral(k*lambda*y/(x^2+y^2)^3/2)*dx)=k*lambda*y*x/(y^2*sqrt(x^2+y^2)) evaluated between -L/2 through L/2 or by multiplying the integral by 2 and then evaluating between 0 and L/2.
The potential is much easier since the potential is a scalar and there is no need to multiply by the cosine of the angle:
Again, multiply by 2 and evaluating between 0 and L/2.
Not very pretty!  


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James J. Kovalcin


I am teaching or have taught AP physics B and C [calculus based mechanics & electricity and magnetism] as well as Lab Physics for college bound students. I have a BS in Physics from the University of Pittsburgh and a Master of Arts in Teaching from same. I have been teaching physics for 34 years. I am constantly updating my skills and have a particular interest in modern physics topics.

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