Respuesta :
Answer:
A) E = 3.70*10^{4} N/C
B) E = 2.281*10^3 N/C
Explanation:
given data:
charge density [tex] \lambda = 130*10^{-9} C/m[/tex]
length of wire = 9.50 cm
a) at x = 4.5 m above midpoint, electric field is calculated as
[tex]E = \frac{1}{ 2\pi \epsilon} * \frac{ \lambda}{x\sqrt{(x^2/a^2)+1}}[/tex]
x = 4.5 cm
midpoint a = 4.5 cm = 0.0475 m
[tex]E =2{\frac{1}{ 8.99*10^9} * \frac{130*10^{-9} }{0.045\sqrt{(4.5^2/4.75^2)+1}}[/tex]
E = 3.70*10^{4} N/C
B) when wire is in circle form
[tex]Q = \lambda * L[/tex]
[tex]= 130*10^{-9} *9.5*10^{-2}[/tex]
= 1.235*10^{-8} C
Radius of circle
[tex]r = \frac{L}{2\pi}[/tex]
[tex]r = \frac{9.5*10^{-2}}{2\pi}[/tex]
r = 1.511*10^{-2} m
[tex]E = \frac{1}{ 2\pi \epsilon} * \frac{Qx}{(x^2+r^2)^{3/2}}[/tex]
[tex]E =8.99*10^{9} * \frac{1.23*10^{-8}*4.5*10^{-2}}{((4.5*10^{-2})^2+(1.511*10^{-2})^2)^{3/2}}[/tex]
E = 2.281*10^3 N/C
Part A. The magnitude of the electric field at the midpoint is [tex]18.9 \times 10^3[/tex] N/C.
Part B. When the shape of the wire is a circle then the electric field intensity is [tex]47.17\times 10^3[/tex] N/C.
Electric Field Intensity
Given that, the length of the wire is 9.50 cm and the charge density is 130 nC/m.
Part A.
The electric field at the midpoint of the wire is
[tex]E = \dfrac {q}{4\pi \epsilon} \times \dfrac {1}{z\sqrt{(z^2/a^2+1)}}[/tex]
Where E is the electric field intensity, [tex]\epsilon[/tex] is a permittivity that is [tex]8.85\times 10^{(-12)}[/tex], z is 4.5 cm and a is the midpoint that is 9.5/2 cm and q is the charge density.
[tex]E = \dfrac {130\times 10^-9} {4\times3.14\times 8.85\times 10^{-12}} \times \dfrac {1}{0.045 \sqrt{{0.045}^2 / {0.0475}^2 +1}}[/tex]
[tex]E = 18.9 \times 10^3 \;\rm N/C[/tex]
The magnitude of the electric field intensity is [tex]18.9\times 10^3 \;\rm N/C[/tex].
Part B.
When the wire is a circle then, charge density at the wire is [tex]Q = q \times l[/tex][tex]Q = 130 \times 10^{-9} \times 0.095[/tex]
[tex]Q= 1.235 \times 10^{-8} C[/tex]
And the radius can be calculated as given below.
[tex]r = \dfrac {l}{2\pi}[/tex]
[tex]r = \dfrac {9.50 }{2 \times 3.14}[/tex]
[tex]r = 1.512 \;\rm cm = 0.015 \;\rm m\\[/tex]
The electric field density at the midpoint of the circle is,
[tex]E = \dfrac {Q} {4 \pi \epsilon} \times \dfrac{ z} { (z^2 +r^2)^{3/2}}[/tex][tex]E = \dfrac {1.235 \times 10^{-8} } {4\times 3.14\times 8.85 \times 10^{-12}} \times \dfrac {0.045}{((0.045)^2 + (0.015)^2)^{1.5}}[/tex]
[tex]E = 47.17 \times 10^3 \;\rm N/C[/tex]
The electric field intensity at the midpoint of the circle is [tex]47.17\times10^3 \;\rm N/C[/tex].
For more details about electric field intensity, follow the link given below.
https://brainly.com/question/6779348.
