A conducting sphere of radius 10 cm has an unknown charge. If the electric field 20 cm from the centre of the sphere is 1.5 x 10³ N/C and points radially inward, what is the net charge on the sphere?

Given,

$R\mathrm{adius} \mathrm{of} \mathrm{conducting} \mathrm{sphere}, r = 10 \mathrm{cm} = 0.1 \mathrm{m}$
$\mathrm{Electric} \mathrm{field} , \mathrm{E}=1.5 \times {10}^3\mathrm{N}/\mathrm{C} \mathrm{at} \mathrm{distance} , \mathrm{d} =20 cm= 0.2 \mathrm{m}$

As,                $\boxed{\mathrm{E} = \frac{1}{4{\mathrm{\pi \epsilon }}_0}\frac{\mathrm{q}}{{\mathrm{r}}^2}}$

$\Rightarrow$                  $\mathrm{q} = \mathrm{E}. 4\mathrm{\pi } {\mathrm{\epsilon }}_0. {\mathrm{r}}^2$                       

$\mathrm{q} = 1.5 \times {10}^3 \times 4\mathrm{\pi } \times 8.854\times 10{ }^{-12}\times (0.2{)}^2$

                        $\mathrm{q} = 6.67 \times {10}^{-9}\mathrm{C}$

Here, since electric field is directed radially inward charge q is negative.

Thus,               $$\begin{gathered} \mathrm{q} = - 6.67 \times {10}^{-9}\mathrm{C} \\ = -6.67 \mathrm{nC}. \end{gathered}$$

Given, Φ = –1.0 x 10³ Nm²/C
r₁ = 0.1 m, r₂ = 0.2 m
(a) Doubling the radius of Gaussian surface will not affect the electric flux since the charge enclosed is the same in the two cases. Thus, the flux will remain be the same i.e., –1.0 x 10³ Nm²/C
(b)

               

(a) To the left of the plates, electric fields are equal and opposite as plates are close to each other electric field is zero as surface charge density in outer side is zero. 

(b) To the right of the plates, electric fields are equal and opposite as plates are close to each other electric field is zero. 

(c) Electric fields between the plates are in same direction as total E.F. on both sides of plate due to $\mathrm{\sigma }$ surface charge density = $\frac{\mathrm{\sigma }}{{\mathrm{\epsilon }}_0}$
So, electric field of inner side of plate  = $\frac{\mathrm{\sigma }}{2{\mathrm{\epsilon }}_0}$
and for both plate   $\mathrm{E} = \frac{\mathrm{\sigma }}{2{\mathrm{\epsilon }}_0}+\frac{\mathrm{\sigma }}{2{\mathrm{\epsilon }}_0}$
                              $\mathrm{E} = \frac{\mathrm{\sigma }}{{\mathrm{\epsilon }}_0} = \mathrm{\sigma } \times 4\mathrm{\pi } \times 9 \times {10}^9$              $\mathrm{E} = 17.0 \times {10}^{-22}\times 4 \times 3.14 \times 9 \times {10}^9$ 

$$\begin{gathered} \mathrm{E} = 1921.7 \times {10}^{-13} \\ = 1.92 \times {10}^{-10}\mathrm{N}/\mathrm{C}. \end{gathered}$$