Electromagnetics Test 1

The electric field is del of the potential i.e.

E = -∇V

\(= - \left( {\frac{\partial }{{\partial x}}V\hat x + \frac{\partial }{{\partial y}}V\hat y + \frac{\partial }{{\partial z}}V\hat z} \right)\)

\(\begin{array}{l} = - \left( {6xy\hat x + \left( {3{x^2} - z} \right)\hat y - y\hat z} \right)\\ \therefore {E_{\left( {2, - 1,4} \right)}} = 12\hat x - 8\hat y - \hat z \end{array}\)

Concept:

According to Gauss law net outward electric flux from any closed surface is equal to the total charge enclosed by the volume.

i.e., ψ = Q_enc

For the given charge density, the above can be written as:

ψ = ∫ ρ_v dv

Calculation:

The charge density is given as:

\({\rho _v} = \left\{ {\begin{array}{*{20}{c}} {\frac{1}{{{r^2}}}\frac{C}{{{m^2}}}:0 < r < 3}\\ {0:~r > 3m} \end{array}} \right.\)

For r = 1, the net flux will be:

ψ = ∫ ρv dv   ---(1)

A small differential volume of a sphere is given by:

\(dS={r^2}\sin \theta drd\theta d\phi \)

The net flux using Equation (1) will now be:

\( = \mathop \smallint \limits_{r = 0}^1 \mathop \smallint \limits_{\phi = 0}^{2\pi } \mathop \smallint \limits_{\theta = 0}^\pi \left( {\frac{1}{{{r^2}}}} \right)({r^2}\sin \theta drd\theta d\phi )\)

= 1 × 2 × 2π = 4π C (Option 2 is correct)

As we have already determined that the total electric flux crossing the surface is 1 m, the electric flux density D at r = 1 m will be evaluated as:

Total electric flux ψ = ∮ D ⋅ dS

Since ψ = 4π, we can write:

∮ D ⋅ dS = 4π

D(4πr²) = 4π

\(D = \frac{1}{{{r^2}}} = 1C/{m^2}\)

Thus D = a_r C/m²  (Option 4 is correct)