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debeloglava
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Homework Statement
A uniform line charge of linear charge density [tex]\lambda[/tex] = 5.00 nC/m extends from x = 0 to x = 10m. The magnitude of the electric field at the point y = 12m on the perpendicular bisector of the finite line charge is?
Homework Equations
E = [tex]\int[/tex]dE = [tex]\int[/tex]k(dq) / r2
Ex = [tex]\int[/tex]k(dq)cos[tex]\theta[/tex] / r2
Ey = [tex]\int[/tex]k(dq)sin[tex]\theta[/tex] / r2
dq = [tex]\lambda[/tex](dx)
The Attempt at a Solution
The perpendicular bisector of the finite line charge occurs at x =5m. I have placed a point (P) at the coordinates on the xy plane at P = (5, 12). With P at this point, the radius from (0, 0) to P is:
r = [tex]\sqrt{(5^2) + (12^2)}[/tex] = [tex]\sqrt{169}[/tex] = 13
If 'b' is the side of the Pythagorean triple equal to 12m then sin[tex]\theta[/tex] = b / r = 12 / 13
With point P bisecting the finite line charge along the x-axis from 0 to 10m I'm arguing by symmetry that Ex = 0, thus there is only a y-component to the electric field.
I'm new to LaTeX and the forums and can't quite seem to get the integration limits to appear properly so the integral is from 0 to 5.
Ey = [tex]\int[/tex]k(dq)sin[tex]\theta[/tex] / r2
Ey = [tex]\int[/tex]k([tex]\lambda[/tex]dx) / r2
Ey = [tex]\int[/tex]k([tex]\lambda[/tex]dx)(12/13) / (132)
Ey = k[tex]\lambda(12/13) / (132) * [/tex][tex]\int[/tex]dx
Ey = 12k[tex]\lambda[/tex] / (133) * [5 - 0]
Ey = (9 x 109)(5 x 10-9)(12 * 5) / (133)
Ey = 1.23 N/C
Using the idea of symmetry again, E = 2 * Ey
E = 2 * (1.23 N/C)
E = 2.46 N/C
Is my work/thought process and ultimately answer correct?