Respuesta :
Answer:
Explanation:
Given that,
Length of wire = 9.5 cm =
Charge density = 100 NC/m
Distance = 4.50 cm
(A). We need to calculate the the magnitude and direction of the electric field
Using formula of magnetic field
[tex]E=\dfrac{k\lambda\times l}{x\sqrt{x^2+a^2}}[/tex]
Here, x = distance from mid point of the wire
Put the value into the formula
[tex]E=\dfrac{9\times10^{9}\times100\times10^{-9}\times9.5\times10^{-2}}{4.50\times10^{-2}\sqrt{(4.50)^2+(\dfrac{9.5\times10^{-2}}{2})^2}}[/tex]
[tex]E=2.90\times10^{4}\ N/C[/tex]
The electric field is [tex]2.90\times10^{4}\ N/C[/tex].
The direction of electric field is upward.
(B). If the wire is now bent into a circle lying flat on the table,
We need to calculate the the magnitude and direction of the electric field
Using formula of electric field
[tex]E=\dfrac{k\lambda lx}{(x^2+(\dfrac{L}{\2\pi})^2)^{\dfrac{3}{2}}}[/tex]
Put the value into the formula
[tex]E=\dfrac{9\times10^{9}\times100\times10^{-9}\times9.5\times10^{-2}\times4.50\times10^{-2}}{((4.50\times10^{-2})^2+(\dfrac{9.5\times10^{-2}}{2\pi})^2)^{\dfrac{3}{2}}}[/tex]
[tex]E=3.59\times10^{4}\ N/C[/tex]
The electric field is [tex]3.59\times10^{4}\ N/C[/tex]
The direction of electric field is upward.
Hence, This is the required solution.
