Applying Gauss' Law: Solving Electric Field Problems

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In summary, the conversation discussed the use of the equation ∮E.dA = qencl/ϵ0 to determine the magnitude and direction of the electric field at different radial distances from a charged cylinder. The charge on the interior and exterior surfaces of the cylinder were also calculated.
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yohanes vianei
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Homework Statement


upload_2019-2-16_22-26-27.png


Homework Equations


E.dA = qencl/ϵ0

The Attempt at a Solution


(a) magnitude E

we use∮E.dA = qencl/ϵ0For a cylinder:∮E.dA = E(2πrL), thenqencl/ϵ0 = E(2πrL)E = qencl/(2πrLϵ0) E = λ/(2πrϵ0); WITH λ = qencl/L



(a) the magnitude E, qencl = Q1 + Q2 , thenE = (Q1 + Q2)/(2πrLϵ0) Because:

· Q1 + Q2 = +3.40 x 10─12 C +( ─2 x 3.40 x 10─12 C) = ─3.40 x 10─12 C,

· Radius, r = 2.00R2 = 2.00 (10R1) = 20 x 1.3 mm = 26 x 10─3 m

· length L = 11.00 m

then,E = (─3.40 x 10─12 C)/(2π x 26 x 10─3 m x 11.00 m x 8.85 x10-12 C2/Nm2)

E = 0.214 N/C(b) direction (radially inward or outward) of the electric field at radial distance r = 2.00R2 is radially inward since qencl < 0(c) E for r = 5.00R1 = 5.00 x 1.3 mm = 6.5 x 10─3 m, and qencl = Q1 = +3.40 x 10─12 C, thenE = (+3.40 x 10─12 C)/(2π x 6.5 x 10─3 m x 11.00 m x 8.85 x10-12 C2/Nm2)

E = 0.855 N/C(d) the direction at r = 5.00R1 is radially outward since qencl > 0(e) the charge on the interior isQinner = ─ Q1 = ─3.40 x 10─12 C(f) the charge on the exterior surface of the shell is

Qouter = Q2 ─ Qinner = (─2 x 3.40 x 10─12 C) ─ (─3.40 x 10─12 C)

Qouter = ─3.40 x 10─12 C[/B]
 

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Hi johanes vianei and welcome to PF.

Is there a question you wish to ask?
 
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Related to Applying Gauss' Law: Solving Electric Field Problems

1. What is Gauss' Law?

Gauss' Law is a fundamental principle in electromagnetism that relates the electric field at a point to the charge enclosed by a closed surface surrounding that point.

2. How is Gauss' Law used to solve electric field problems?

Gauss' Law allows us to calculate the electric field at a point due to a distribution of charges by integrating over a closed surface surrounding that point. This simplifies the calculation of the electric field compared to using Coulomb's Law for each individual charge.

3. What is the equation for Gauss' Law?

The equation for Gauss' Law is ∮SE⋅dA = Qenc0, where ∮SE⋅dA is the flux of the electric field over a closed surface S, Qenc is the total charge enclosed by S, and ε0 is the permittivity of free space.

4. What are some common applications of Gauss' Law?

Gauss' Law is used in many areas of physics and engineering, including designing circuits, analyzing electric fields in materials, and understanding the behavior of charged particles. It is also used in practical applications such as calculating the electric field inside a capacitor or in a parallel plate capacitor.

5. Are there any limitations to using Gauss' Law?

While Gauss' Law is a powerful tool for solving electric field problems, it does have some limitations. It can only be used for problems with high symmetry, such as spherical, cylindrical, or planar symmetry. Additionally, it assumes a static electric field and does not take into account the effects of changing magnetic fields.

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