Respuesta :
Answer:
[tex]E=\dfrac{\lambda }{2\pi \varepsilon _or}[/tex]
Explanation:
Given that
For straight wire
Charge density= λ
For hollow metal cylinder
Charge density=2 λ
We know that electric filed for wire given as
[tex]E_w=\dfrac{\lambda_{wire} }{2\pi \varepsilon _or}[/tex]
[tex]E_w=\dfrac{\lambda }{2\pi \varepsilon _or}[/tex]
Now the electric filed due to hollow metal cylinder
[tex]E_c=\dfrac{\lambda_{cylinder} }{2\pi \varepsilon _or}[/tex]
[tex]E_c=\dfrac{2\lambda }{2\pi \varepsilon _or}[/tex]
Now by considering the Gaussian surface r<R then only electric fild due to wire will present.So
At r<R
[tex]E=\dfrac{\lambda }{2\pi \varepsilon _or}[/tex]
The expressions for the magnitude of the electric field strength inside the cylinder, r is,
[tex]E=\dfrac{\lambda }{2\varepsilon_o \pi r}\\[/tex]
What is electric field?
The electric field is the field, which is surrounded by the electric charged. The electric field is the electric force per unit charge.
Here a long, thin straight wire with linear charge density λ runs down the center of a thin, hollow metal cylinder of radius R.
As the electric flux is the product of electric flux and the area of the cylinder. Thus,
[tex]\phi =E(2\pi rL)[/tex] ......1
This electric flux can also be given as,
[tex]\phi =\dfrac{\lambda L}{\varepsilon_o}[/tex] ......2
Where, ([tex]\lambda[/tex]) is the linear charge density. On comparing both the equation, we get,
[tex]E(2\pi rL)=\dfrac{\lambda L}{\varepsilon_o}\\E=\dfrac{\lambda L}{\varepsilon_o (2\pi rL)}\\E=\dfrac{\lambda }{2\varepsilon_o \pi r}\\[/tex]
The cylinder has a net linear charge density 2λ where λ is positive. Thus for this hollow cylinder, the electric filed can be given as,
[tex]E=\dfrac{2\lambda }{2\varepsilon_o \pi r}\\[/tex]
For charge inside the cylinder, where r < R, can be expressed with the following expression,
[tex]E=\dfrac{\lambda }{2\varepsilon_o \pi r}\\[/tex]
Learn more about electric field here;
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