Poll Results
No votes. Be the first one to vote.
Lost your password? Please enter your email address. You will receive a link and will create a new password via email.
Please briefly explain why you feel this question should be reported.
Please briefly explain why you feel this answer should be reported.
Please briefly explain why you feel this user should be reported.
To find the electric flux density ( D ) at ( R = 6m ) due to the three charged cylindrical sheets, we can use Gauss’s Law for electric fields. The key steps are:
1. Identify the charge density: The charge densities given are:
– For the first sheet at ( R = 2m ), ( sigma_1 = 5 , text{C/m}^2 )
– For the second sheet at ( R = 4m ), ( sigma_2 = -2 , text{C/m}^2 )
– For the third sheet at ( R = 5m ), ( sigma_3 = -3 , text{C/m}^2 )
2. Calculate the contribution to the flux density ( D ) at ( R = 6m ) from each sheet. The formula for the electric flux density ( D ) due to an infinite plane sheet of charge is:
[
D = frac{sigma}{2} quad text{(points away from positively charged sheet)}
]
– For the first sheet:
[
D_1 = frac{5}{2} = 2.5 , text{C/m}^2 quad (text{at } R > 2m)
]
– For the second sheet:
Answer: d
Explanation: The radius R = 6m encloses all the three Gaussian cylinders.
By Gauss law, ψ = Q
D(2πRL) = σ(2πRL), D(2π X 6) = Q1 + Q2 + Q3 = σ1(2π X 2) + σ2(2π X 4) + σ3(2π X
5), here σ1 = 5, σ2 = -2 and σ3 = -3. We get D = -13/6 units.