91Ó°ÊÓ

The radius of both disks is, $R$.

The uniform charge carried by disk $A$is, $Q_A$.

The uniform charge carried by disk $B$is,$Q_B$.

The distance between the plates is, $d=0.1\text{mm}$.

The potential difference between the plates is,$V_B-V_A=-10\text{V}$.

When a charge moves from one location to other in an electric field then the potential difference between two locations changes with the travel distance.

If the distance traveled by a charge increases then the value of the potential difference between locations also increases.

The formula for the potential difference from disk A to disk B is given by,

$$\begin{gathered} V_B-V_A=\frac{1}{4\mathrm{Ï€}{\mathrm{Î\mu }}_0}\frac{Q_B}{R}-\frac{1}{4\mathrm{Ï€}{\mathrm{Î\mu }}_0}\frac{Q_A}{R} \\ {V}_B-V_A=\frac{1}{4\mathrm{Ï€}{\mathrm{Î\mu }}_0}\frac{\left(Q_B-Q_A\right)}{R} \end{gathered}$$

Here, $\frac{1}{4\mathrm{Ï€}{\mathrm{Î\mu }}_0}$is a constant and its value is $9×{10}^9\text{N}·{\text{m}}^2{\text{/C}}^2$.

$\begin{array}{rcl}-10\text{V}×\frac{1\text{N}·\text{m/C}}{1\text{V}}& =& \left(9×{10}^9\text{N}·{\text{m}}^2{\text{/C}}^2\right)\frac{\left(Q_B-Q_A\right)}{R}\\ \frac{\left(Q_B-Q_A\right)}{R}& =& \frac{-10\text{N}·\text{m/C}}{\left(9×{10}^9\text{N}·{\text{m}}^2{\text{/C}}^2\right)}\\ \frac{\left(Q_B-Q_A\right)}{R}& =& \frac{-10}{\left(9×{10}^9\right)}\text{C/m}\end{array}$

Assuming,$Q_B=10\text{C}$, $Q_A=20\text{C}$and $R=9×{10}^9\text{m}$,

$\begin{array}{rcl}\frac{\left(10\text{C}-20\text{C}\right)}{9×{10}^9\text{m}}& =& \frac{-10}{\left(9×{10}^9\right)}\text{C/m}\\ \frac{-10}{\left(9×{10}^9\right)}\text{C/m}& =& \frac{-10}{\left(9×{10}^9\right)}\text{C/m}\\ \text{LHS}& =& \text{RHS}\end{array}$

Hence, the possible values are $Q_B=10\text{C}$, $Q_A=20\text{C}$and $R=9×{10}^9\text{m}$.