Electric field at a point from an insulating shell

In summary, an infinite line of charge with linear density λ1 = 6.9μ C/m is surrounded by a thick insulating shell with inner radius a = 2.6 cm and outer radius b = 4.8 cm. The insulating shell has a volume density of ρ = -656μ C/m3 and a linear charge density of λ2 = -3.36E-6 C/m. To calculate the y-component of the electric field at point P, located 8.3 cm along the y-axis from the line of charge, Gauss' Law (equation 2) must be applied using the charge enclosed by the Gaussian surface.
  • #1
rezal
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Homework Statement


An infinite line of charge with linear density λ1 = 6.9μ C/m is positioned along the axis of a thick insulating shell of inner radius a = 2.6 cm and outer radius b = 4.8 cm. The insulating shell is uniformly charged with a volume density of ρ = -656μ C/m3
h4_cylinder.png

1)What is λ2, the linear charge density of the insulating shell?

2)What is Ey(P), the value of the y-component of the electric field at point P, located a distance 8.3 cm along the y-axis from the line of charge?

Homework Equations


[tex] 1) E=\frac{1}{4\pi\epsilon_0} \frac{q_0}{r^2} [/tex]
[tex] 2) \int E \bullet dA = \frac{Q}{\epsilon_0} [/tex][/B]

The Attempt at a Solution


For 1, I found λ2 with [tex] \rho\pi(b^2 - a^2) [/tex] and it turned out to be λ2 = -3.36E-6 C/m

I'm stuck on 2, I tried doing it by adding three electric fields. The first one in the shells' center using λ1 with equation 1, the second one in the shell using the charge from ρV with equation 2, and the third one in a cylindrical gaussian surface that has P on its surface with equation 2. I keep getting 1.8E8 C/m which is wrong.

Can someone tell me where I'm going wrong with this?[/B]
 
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  • #2
rezal said:

Homework Statement


An infinite line of charge with linear density λ1 = 6.9μ C/m is positioned along the axis of a thick insulating shell of inner radius a = 2.6 cm and outer radius b = 4.8 cm. The insulating shell is uniformly charged with a volume density of ρ = -656μ C/m3
h4_cylinder.png

1)What is λ2, the linear charge density of the insulating shell?

2)What is Ey(P), the value of the y-component of the electric field at point P, located a distance 8.3 cm along the y-axis from the line of charge?

Homework Equations


[tex] 1) E=\frac{1}{4\pi\epsilon_0} \frac{q_0}{r^2} [/tex]
[tex] 2) \int E \bullet dA = \frac{Q}{\epsilon_0} [/tex][/B]

The Attempt at a Solution


For 1, I found λ2 with [tex] \rho\pi(b^2 - a^2) [/tex] and it turned out to be λ2 = -3.36E-6 C/m

I'm stuck on 2, I tried doing it by adding three electric fields. The first one in the shells' center using λ1 with equation 1, the second one in the shell using the charge from ρV with equation 2, and the third one in a cylindrical gaussian surface that has P on its surface with equation 2. I keep getting 1.8E8 C/m which is wrong.

Can someone tell me where I'm going wrong with this?[/B]

Equation 1 can not be used, as it refers to a point charge. Apply Gauss' Law (eq. 2 ) and remember, Q is the charge enclosed by the Gaussian surface.
 
  • #3
ehild said:
Equation 1 can not be used, as it refers to a point charge. Apply Gauss' Law (eq. 2 ) and remember, Q is the charge enclosed by the Gaussian surface.
Ah alright, I got it now. Thank you!
 

FAQ: Electric field at a point from an insulating shell

What is an insulating shell?

An insulating shell is a material that does not allow the flow of electric current through it. This means that the electrons in the material are tightly bound and cannot move freely.

How is the electric field at a point determined from an insulating shell?

The electric field at a point from an insulating shell is determined by the amount of charge on the shell and the distance between the point and the shell. The greater the charge and the closer the point is to the shell, the stronger the electric field will be.

What is the relationship between the electric field and the distance from an insulating shell?

The electric field from an insulating shell follows an inverse square law, meaning that as the distance from the shell increases, the electric field decreases proportionally. This relationship is similar to that of gravity.

Can the electric field from an insulating shell be negative?

Yes, the electric field from an insulating shell can be negative. This means that the direction of the electric field is opposite to the direction of the positive charge on the shell. Negative electric fields can also indicate a decrease in the overall strength of the electric field.

How can the electric field from an insulating shell be used in practical applications?

The electric field from an insulating shell has many practical applications, such as in the design of electronic devices and in the field of electrostatics. It also plays a role in the behavior of lightning and in the formation of thunderstorms. Additionally, it can be used in the creation of insulating materials for electrical insulation purposes.

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