Electrostatics

$\frac{kQe}{0.12}+\frac{1}{2}m{u}^2=\frac{kQe}{0.15}+\frac{1}{2}m{u}^2$

$\frac{kQe*0.03}{0.12*0.15}=\frac{1}{2}m*\left(v^2-u^2\right)$

$Q=\frac{9.1*99*10^{-11}}{12}$

$=\frac{75.075*10^{-12}}{4\pi *{\left(0.1\right)}^2}=5.97nC/m^2$

$WD=\Delta QV$

$\Rightarrow \Delta Q=4\mu C$

$Q_i=\frac{2*4}{2+4}*6=8\mu C$

$Q_f=12\mu C=C_{eq}*6$

$C_{eq}=2\mu F\Rightarrow C_1=C_2=4\mu F$

$k=\frac{C_1}{C_0}=2$

$Q_1=C_{1}V=2V\mu C$

$Q_2=Q_3=\frac{C_2{C}_3}{C_2+C_3}V=\frac{6*12}{6+12}V=4V\mu C$

$Q_1:Q_2:Q_3=1:2:2$

V = 10 $\left(1?e^{?t/RC}\right)$ ${}^{}$

2 = 20 $\left(1?e^{?t/RC}\right)$ ${}^{}$

$\frac{1}{10}=1?e^{?t/RC}$

$e^{t/RC}=\frac{10}{9}$

$\frac{t}{RC}=ln\left(\frac{10}{9}\right)=0.105$

$C=\frac{t}{R*0.105}=\frac{10^{?6}}{10*0.105}=0.95?F$

V_common = (C? V? + C? V? )/ (C? + C? )? 2 = (3? F)*12)/ (3? F)+ (3? F)*K)? 2 = 12/ (K+1)? K = 5

tan 60° = (2kλ? /y? )/ (2kλ? /x? )

(√3λ? /x? ) = (λ? /y? ); √3λ? y? = λ? x?

$V_1=\frac{k{Q}_3}{3R}+\frac{k{Q}_2}{2R}+\frac{k{Q}_1}{R}=30$

$V_2=\frac{k{Q}_3}{3R}+\frac{k{Q}_2}{2R}+\frac{k{Q}_1}{2R}=15$

$V_3=\frac{k{Q}_3}{3R}+\frac{k{Q}_2}{3R}+\frac{k{Q}_1}{3R}=10$

Eq. (1) $\&$  (2) $\to \frac{k{Q}_1}{R}=30$  

Eq. (2) $\&$  (3) $\to \frac{k{Q}_2}{6R}+\frac{k{Q}_1}{6R}=5\Rightarrow \frac{k{Q}_2}{R}=-\frac{k{Q}_1}{R}+30=0$

By (3) $\to \frac{k{Q}_3}{R}=30-\frac{k{Q}_2}{R}-\frac{k{Q}_1}{R}=0$

Potential of innermost shell (earthed) $=\frac{k{Q}_1^{\mathrm{\text{'}}}}{R}+\frac{k{Q}_2}{2R}+\frac{k{Q}_3}{3R}=0$

$\Rightarrow \mathrm{}\frac{k{Q}_1^{\mathrm{\text{'}}}}{R}=0\mathrm{}\therefore \mathrm{}{Q}_1^{\mathrm{\text{'}}}=0$

Charge initially on capacitor 1=10µC
Charge initially on capacitor 2=20µC
Finally the potential difference across both capacitor will be equal.
Let us assume that to be V.
1V + 2V = +10
V = 10 / 3 = 3.33V

If charge (-Q) at origin is replaced by (+Q), then electric field at the centre of the cube is zero. Thus, electric field at the centre of the cube is as if only (-2Q) charge is present at the origin.

$\overrightarrow{E}=\frac{1}{4\pi {\epsilon }_0}\frac{\left(-2Q\right)}{{\left(\frac{\sqrt[]{3}}{2}a\right)}^2}.\frac{1}{\sqrt[]{3}}\left(\widehat{x}+\widehat{y}+\widehat{z}\right)=\frac{|-2Q}{3\sqrt[]{3}\pi {\epsilon }_0{a}^2}\left(\widehat{x}+\widehat{y}+\widehat{z}\right)$