Question

Charge of uniform surface density $\mathbf{8}\mathbf{.}\mathbf{00}\mathbf{}\mathbf{nC}\mathbf{/}{\mathbf{m}}^{\mathbf{2}}$is distributed over an entire x-yplane; charge of uniform surface densityis $\mathbf{3}\mathbf{.}\mathbf{00}\mathbf{}\mathbf{nC}\mathbf{/}{\mathbf{m}}^{\mathbf{2}}$distributed over the parallel plane defined by z = 2.00 m. Determine the magnitude of the electric field at any point having a z-coordinate of

(a)1.00 m and

(b) 3.00 m.

Short answer

1. The magnitude of the electric field at any point having a z-coordinate of 1m is $2.82\times {10}^2\mathrm{N}/\mathrm{C}$.
2. The magnitude of the electric field at any point having a z-coordinate of 3m is $6.21\times {10}^2\mathrm{N}/\mathrm{C}$.

Step-by-step solution

The given data

1. Surface charge density $8.00\mathrm{nC}/{\mathrm{m}}^2$is distributed over an entire x-y plane.
2. Surface charge density $3.00\mathrm{nC}/{\mathrm{m}}^2$is distributed over the parallel plane defined by: z = 2.00 m

Understanding the concept of the electric field

Using the concept of the electric field of a Gaussian surface, we can calculate the electric fields at the given z-coordinates. The net electric field, due to the presence of two surface charge densities σ, is used for the net electric field.

Formula:

The electric field of a non-conduction sheet,

$E=\frac{\sigma }{2{\epsilon }_0}$ (i)

a) Calculation of the electric field having z = 1 m

Both sheets are horizontal (parallel to the x-y plane), producing vertical fields (parallel to the z-axis). At points above the z = 0 sheet (sheet A), its field points upward (toward +z); at points above the z = 2 sheet (sheet B), its field does likewise. However, below the z = 2 sheet, its field is oriented downward.

The magnitude of the net field in the region between the sheets is given using equation (i) as follows:

$$\begin{gathered} \left|E\right|=\frac{{\sigma }_A}{2{\epsilon }_0}-\frac{{{\sigma }_A}^{\text{'}}}{2{\epsilon }_0} \\ =\frac{8.00\times {10}^{-9}\mathrm{C}/{\mathrm{m}}^2-3.00\times {10}^{-9}\mathrm{C}/{\mathrm{m}}^2}{2(8.85\times {10}^{-12}{\mathrm{C}}^2/{\mathrm{Nm}}^2)} \\ =2.82\times {10}^2\mathrm{N}/\mathrm{C} \end{gathered}$$

Hence, the value of the electric field is $2.82\times {10}^2\mathrm{N}/\mathrm{C}$.

b) Calculation of the electric field having z = 3 m

The magnitude of the net field at points above both sheets is given using equation (i) as follows:

$$\begin{gathered} \left|E\right|=\frac{{\sigma }_A}{2{\epsilon }_0}+\frac{{{\sigma }_A}^{\text{'}}}{2{\epsilon }_0} \\ =\frac{8.00\times {10}^{-9}\mathrm{C}/{\mathrm{m}}^2-3.00\times {10}^{-9}\mathrm{C}/{\mathrm{m}}^2}{2(8.85\times {10}^{-12}{\mathrm{C}}^2/{\mathrm{Nm}}^2)} \\ =6.21\times {10}^2\mathrm{N}/\mathrm{C} \end{gathered}$$

Hence, the value of the electric field is $6.21\times {10}^2\mathrm{N}/\mathrm{C}$.