Physics Electrostatic Potential and Capacitance

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

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

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

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

ceq = $\frac{24*8}{32}=6\mu F$

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.

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

When connected in series, equivalent capacitance,

$C_1=\frac{C*C}{C+C}=\frac{C}{2}$          

When connected in parallel, equivalent capacitance

C₂ = C + C = 2C

$\frac{C_1}{C_2}=\frac{C/2}{2C}=\frac{1}{4}$            

Ui = ½C (V/3)² + ½C (V/3)² + ½C (2V/3)²
Uf = ½CV²
Wb = CV²/3
ΔH = [Wb - (Uf - Ui)] = CV²/6 = 0.30 mJ

(4/3)πR³ = 27 (4/3)πr³) ⇒ R = 3r (i)
v = Kq/r = V? /r = (q? /q? ) (r? /r? ) (i)
⇒ 220/V? = (q/ (27q) (3r/r) (i)
⇒ 220/V? = 1/9 (i)
⇒ V? = 220 * 9 = 1980 volt (i)

W_agent + W_battery = ΔU
W = ½ (KC - C)V² - [ (KC - C)V]V
= ½ (K - 1)CV²

${\mathrm{V}}_{\mathrm{P}}={\mathrm{V}}_{\mathrm{q}}+{\mathrm{V}}_{-\mathrm{q}}$

$\begin{array}{rr}& =\left(\frac{\mathrm{K}\mathrm{q}}{5-3}+\frac{\mathrm{K}(-\mathrm{q})}{5+3}\right)*{10}^2\\ & =\left(\frac{\mathrm{K}\mathrm{q}}{2}-\frac{\mathrm{K}\mathrm{q}}{8}\right)*{10}^2\\ & =\left(\frac{3\mathrm{K}\mathrm{q}}{8}\right)*{10}^2\end{array}$