91Ó°ÊÓ

a) Three plastic sheets are uniformly charged.

b) The scale of the vertical axis, ${\mathrm{E}}_{\mathrm{s}}=6.0×{10}^5\mathrm{N}/\mathrm{C}$

Using the concept and the given data, we can get the net electric field of the required sheets. Now, using the concept of Gauss's flux theorem, we can get the surface density values of the sheets. Them dividing them, we can get the required ratio.

Formula:

The electric field of a non-conducting sheet,$E=\frac{\mathrm{σ}}{2{\mathrm{Î\mu }}_0}$ (1)

In the region between sheets 1 and 2, the net field is given by:

$E_1-E_2+E_3=2.0×{10}^{5}N/C.....................\left(2\right)$

In the region between sheets 2 and 3, the net field is at its greatest value which is given by:

${\mathrm{E}}_1-{\mathrm{E}}_2+{\mathrm{E}}_3=6.0×{10}^5\mathrm{N}/\mathrm{C}.....................\left(2\right)$

The net field vanishes in the region to the right of sheet 3, where the relation of the electric fields is given as:

$E_1+E_2=E_3.......................\left(4\right)$

So solving equations (2), (3), and (4), we get the values of the electric fields as given:

$$\begin{gathered} {\mathrm{E}}_1=1.0×{10}^5\mathrm{N}/\mathrm{C}; \\ {\mathrm{E}}_2=2.0×{10}^5\mathrm{N}/\mathrm{C}; \\ {\mathrm{E}}_3=3.0×{10}^5\mathrm{N}/\mathrm{C}; \end{gathered}$$

From, equation (1), we can get that the surface density on a sheet is directly proportional to the value of the electric field. Thus, the ratio of the surface densities on sheet 3 to that on sheet 2 is given aslocalid="1657352971674" $\frac{\left|{\mathrm{σ}}_3\right|}{\left|{\mathrm{σ}}_3\right|}=\frac{3.0×{10}^5\mathrm{N}/\mathrm{C}}{2.0×{10}^5\mathrm{N}/\mathrm{C}}$

Hence, the value of the ratio is 1.5.