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

At vero eos et accusamus et iusto odio dignissimos ducimus qui blanditiis praesentium voluptatum deleniti atque corrupti quos dolores et quas molestias excepturi sint occaecati cupiditate non provident, similique sunt in culpa qui officia deserunt mollitia animi, id est laborum et dolorum fuga.
Et harum quidem rerum facilis est et expedita distinctio. Nam libero tempore, cum soluta nobis est eligendi optio cumque nihil impedit quo minus id quod maxime placeat facere possimus, omnis voluptas assumenda est, omnis dolor repellendus.
Itaque earum rerum hic tenetur a sapiente delectus, ut aut reiciendis voluptatibus maiores alias consequatur aut perferendis doloribus asperiores repellat.

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

Hey! We 've verified this expert answer for you, click below to unlock the details :)

SOLVED

At vero eos et accusamus et iusto odio dignissimos ducimus qui blanditiis praesentium voluptatum deleniti atque corrupti quos dolores et quas molestias excepturi sint occaecati cupiditate non provident, similique sunt in culpa qui officia deserunt mollitia animi, id est laborum et dolorum fuga.
Et harum quidem rerum facilis est et expedita distinctio. Nam libero tempore, cum soluta nobis est eligendi optio cumque nihil impedit quo minus id quod maxime placeat facere possimus, omnis voluptas assumenda est, omnis dolor repellendus.
Itaque earum rerum hic tenetur a sapiente delectus, ut aut reiciendis voluptatibus maiores alias consequatur aut perferendis doloribus asperiores repellat.

I got my questions answered at brainly.com in under 10 minutes. Go to brainly.com now for free help!

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

\[E = \frac {1}{4.\pi.\epsilon _{0}} . 2. \int\limits_{0}^{\infty} \frac {dq}{r²} = \frac {1}{4.\pi.\epsilon _{0}} .2.\int\limits_{0}^{\infty} \frac {0,3µC/m.dz}{z²+3²}\]
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)

Not the answer you are looking for? Search for more explanations.

More answers

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

ok I have found the answer, it is D) 5,2kN/C

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.