As V = $${q \over C}$$

$$ \therefore $$ V $$ \propto $$ $${1 \over C}$$

So $${{{V_1}} \over {{V_2}}} = {{{C_2}} \over {{C_1}}}$$

$$ \Rightarrow $$ $${{{V_1}} \over {{V_2}}} = {6 \over 4} = {3 \over 2}$$

Also V₁ + V₂ = 10

$$ \therefore $$ V₁ + $${2 \over 3}$$V₁ = 10

$$ \Rightarrow $$ $${{5{V_1}} \over 3}$$ = 10

$$ \Rightarrow $$ V₁ = 6 V

The charge on 4 $$\mu $$F capacitor is = C₁V₁ = 4 $$ \times $$ 6 = 24 $$\mu $$C