Consider a cylinder charge distribution extending from r=0 to r=9.4 cm of charge

Question

Consider a cylinder charge distribution extending from r=0 to r=9.4 cm of charge density d=r/ao. Where  ao= 22cm*m^3/mc

This cylindrical distribution is serenaded by a dielectric shell of dielectric constant 2.32 whose inner radius is 17.8 cm and outer radius is 24.8cm.

What is the electric field at 4.86cm? Assume the length L is very long compared to the diameter of the dielectric shell and neglect edge effects. The permittivity of a vacuum is 8.8542e-12 C^2/N*m^2

Consider a cylindrical charge distribution extending from r = 0 to r = 9.4 cm of charge density rho = r/a0, where a0 = 22 cm m3/mu C. This cylindrical charge distribution is surrounded by a dielectric shell of dielectric constant 2.32 whose inner radius is 17.8 cm and outer radius is 24.8 cm. What is the electric field at 4.86 cm? Assume the length L is very long compared to the diameter of the dielectric shell and neglect edge effects. The permittivity of a vacuum is 8.8542 times 10-12 C2/N m2. Answer in units of V/m

Explanation / Answer

EF = Lambda/2pie0R

density = charge /volume = 9.4/22. = 0.427 uC/m^3

charge/L= lambda = = 0.427uC * 2*3.14* 0.094*0.094= 0.0554 nC/m

now E = L/2pi e0 R = 0.0554*10^-9/2*3.14*8.85*10^-12* 0.0486

E = 20.5 N/C

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