A point charge of 30 nC is located at the origin, while plane Y = 3 carries charge 10 nC/m². Find flux density at (0, 4, 3). 

Concept:

The electric flux density at a point due to multiple charges is the vector sum of the flux densities due to each charge.

For a point charge at the origin, \( \vec{D} = \frac{Q}{4\pi r^3} \vec{r} \) and for a charged infinite sheet at Y=3, the field is uniform with magnitude \(\vec{D} = \frac{\rho_s}{2} \vec{a_y} \) above or below the sheet.

Calculation:

Given:

\(Q = 30~nC, \rho_s = 10~nC/m^2, point: (0,4,3)\)

Position vector: \( \vec{r} = 0\vec{a_x} +4\vec{a_y} +3\vec{a_z} \), magnitude: r = 5

Field due to point charge:

\( \vec{D}_Q = \frac{30}{4\pi (5)^3} (4\vec{a_y} +3\vec{a_z}) = 0.0764\vec{a_y} +0.0573\vec{a_z} \)

Field due to sheet charge at Y=3:

\( \vec{D}_{sheet} = \frac{\rho_s}{2} \vec{a_y} = 5\vec{a_y} \)

Total field:

\( \vec{D} = \vec{D}_Q + \vec{D}_{sheet} = (5+0.0764)\vec{a_y} +0.0573\vec{a_z} \approx 5.076\vec{a_y} +0.0573\vec{a_z}~nC/m^2 \)

Answer:

The flux density at (0,4,3) is: (5.076 ay + 0.0573 az) nC/m2