anonymous
  • anonymous
An infinite line charge of linear density λ = 0.30 μC/m lies along the z axis and a point charge q = 6.0 µC lies on the y axis at y = 2.0 m. The x component of the electric field at the point P on the x axis at x = 3.0 m is approximately A) 1.8 kN/C B) 4.2 kN/C C) 0.96 kN/C D) 5.2 kN/c E) 0.64 mN/C
Physics
  • Stacey Warren - Expert brainly.com
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SOLVED
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katieb
  • katieb
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fretje
  • fretje
i calculate the field from the point charge q: \[E = \frac{1}{4.\pi.\epsilon _{0}}.\frac {q}{r ^{2}}.r ^{.}\] epsilon is permittivity of free space, r distance to point charge, q is charge, and r point is the unit vector to the point p (q1->P). so \[E = \frac{1}{4.\pi.8,85.10E-12}.\frac {6µC}{3,6^{2}}.r ^{.}\] is E= 4162,87 N/C = 4,16kN/C Logically the E field must be still bigger than this number, since there is still the charges on the Z-axis, (also positive charges). So the answer must be either B or D. Now I must integrate the charges on Z-axis for to obtain the field in P, than add to the previous result.
fretje
  • fretje
\[E = \frac {1}{4.\pi.\epsilon _{0}} . 2. \int\limits_{0}^{\infty} \frac {dq}{r²} = \frac {1}{4.\pi.\epsilon _{0}} .2.\int\limits_{0}^{\infty} \frac {0,3µC/m.dz}{z²+3²}\] which means i have to integrate from z= 0 to z= infinity, since the design is symmetric over the z axis, I multiply by 2 (same result as integrating from -infinity to + infinity), where dq is the charge from an infinitely small point dz on the z axis. also distance of r² = 3²+z² (pythagoras right triangle)
fretje
  • fretje
|dw:1332457234242:dw|

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fretje
  • fretje
ok now I solve the integral: found it on http://en.wikipedia.org/wiki/List_of_integrals_of_rational_functions \[\int\limits_{0}^{\infty} \frac{dz}{z²+3²} = \frac {1}{3}.\tan^{-1} \frac{z}{3}\] for bounderies 0 to infinity \[E = \frac {1}{4.\pi.\epsilon _{0}} .2.\int\limits\limits_{0}^{\infty} \frac {0,3µC/m.dz}{z²+3²}\] becomes then \[E = 5395.\frac {1}{3}\tan^{-1} \frac{z}{3} for z= 0 \rightarrow \infty\] which is \[E = 5395 . \frac {1}{3} . \frac {\pi}{2} = 2823 N/C= 2,8kN/C\] And I have a wrong answer. Sorry about that. It was a rough day. I hope I could help. Tomorrow better.
fretje
  • fretje
ok I have found the answer, it is D) 5,2kN/C
fretje
  • fretje
Here comes the final solution: \[Ex = \frac{1}{4.\pi.\epsilon _{0}}.\frac {q}{r ^{2}}.\cos \theta \] where cos theta is the projection of the E field on the x-axis, to obtain the x component only of the E field. \[\cos \theta = \frac {3m}{3,6m} \] thus i get for the E field from the point charge q \[E = \frac{1}{4.\pi.8,85^{-12}}.\frac {6µC}{3,6^{2}}.0,8333 = 3,469kN/C\] Next the for me tougher part of integrating the charge on the z-axis over the whole z-axis |dw:1332535357784:dw| I make up the relation between an infinitesimal piece dz on z-axis and the point p on the x-axis. \[E = \frac {1}{4.\pi.\epsilon _{0}}.2.0,3^{-6}C/m. \int\limits_{0}^{\infty} \frac{\cos \theta.dz}{z²+3²}\] I substitute cos theta with \[\frac{3}{\sqrt{z²+3²}}\] which is seen from the right triangle and thus get \[E = \frac {1}{4.\pi.\epsilon _{0}}.2.0,3^{-6}C/m. \int\limits_{0}^{\infty} \frac{\frac{3}{\sqrt{z²+3²}}.dz}{z²+3²}\] or worked out \[E = \frac {1}{4.\pi.\epsilon _{0}}.2.0,3^{-6}C/m. \int\limits_{0}^{\infty} \frac{3.dz}{(z²+3²)^{\frac{3}{2}}}\] To solve this integral I did substitution z = a.tan(u) or z² = a² tan²(u) with a = 3 dz is then dz = 3.sec²u.du the integral becomes \[E = \frac {1}{4.\pi.\epsilon _{0}}.2.0,3^{-6}C/m. \int\limits\limits_{0}^{\infty} \frac{3.3.\sec ^{2} u.du}{9.\left( \tan ^{2}u+1 \right)^{\frac {3}{2}}}\] and \[E = \frac {1}{4.\pi.\epsilon _{0}}.2.0,3^{-6}C/m. \int\limits\limits_{0}^{\infty} \frac{3.3.\sec ^{2} u.du}{9.\left( \sec ^{2}u \right)^{\frac {3}{2}}}\] and \[E = \frac {1}{4.\pi.\epsilon _{0}}.2.0,3^{-6}C/m. \int\limits\limits_{0}^{\infty} \frac{3.du}{9. \sec u} \] \[E = \frac {1}{4.\pi.\epsilon _{0}}.2.0,3^{-6}C/m.\frac {1}{3}. \int\limits\limits_{0}^{\infty} \cos u.du\] is \[E = \frac {1}{4.\pi.\epsilon _{0}}.2.0,3^{-6}C/m.\frac {1}{3}.\sin u \] and since \[\sin u = \frac{z}{\sqrt{z²+3²}}\] (see triangle:) |dw:1332539430235:dw| i have \[E = \frac {1}{4.\pi.\epsilon _{0}}.2.0,3^{-6}C/m.\frac {1}{3}.\frac {z}{\sqrt {z²+9} } \] with integration bounderies 0 to infinity The limit \[ \lim_{0 \rightarrow \infty} \frac {z}{\sqrt {z²+9} } \] makes 1 so I have \[E = \frac {1}{4.\pi.\epsilon _{0}}.2.0,3^{-6}C/m.\frac {1}{3} =\[1798N/C +\] \] Now I add the two results: \[3469N/C + 1798N/C = 5267N/C\] Which comes closest to answer D.

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