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 evaluate Gauss’s law using the electric displacement field ( mathbf{D} ), we can use the integral form:
[
Phi_D = iint_S mathbf{D} cdot dmathbf{A} = int_V rho_f dV
]
where ( Phi_D ) is the electric displacement flux, ( S ) is the closed surface surrounding volume ( V ), and ( rho_f ) is the free charge density in the volume.
Given the electric displacement field ( mathbf{D} = frac{5r^2}{4} mathbf{i} ) in spherical coordinates, we can find ( D_r, D_theta, ) and ( D_phi ).
Since in spherical coordinates the unit vectors are ( hat{r}, hat{theta}, hat{phi} ), and our given ( D ) component only has an i component, which corresponds to the x-direction in Cartesian coordinates, we need to express it in spherical coordinates:
[
mathbf{D} = frac{5r^2}{4} hat{i} = frac{5r^2}{4} sintheta cosphi hat{r} + frac{5r^2}{4} sintheta sinphi hat{theta} + frac{5r^2}{4} cos
Answer: b
Explanation: ∫∫ D.ds = ∫∫∫ Div (D) dv, where RHS needs to be computed.
The divergence of D given is, Div(D) = 5r and dv = r2 sin θ dr dθ dφ. On integrating, r =
0->4, φ = 0->2π and θ = 0->π/4, we get Q = 588.9.
Answer: c
Explanation: ∫∫ ( 5r2/4) . (r2 sin θ dθ dφ), which is the integral to be evaluated.
Put r = 4m and substitute θ = 0→ π/4 and φ = 0→ 2π, the integral evaluates to 588.9.