Applying Gauss' Law: Solving Electric Field Problems

[yohanes vianei]

Homework Statement

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]

 

[kuruman]

Hi johanes vianei and welcome to PF.

Is there a question you wish to ask?