Detailed solution not required.
$\epsilon_{0} = \frac{1}{4\pi F}\frac{e^{2}}{r^{2}}$
Due to induction, net charges on the outer surface of the spheres are established as shown in standard diagrams.
$\sigma = \frac{Q_{1}}{4\pi R^{2}} = \frac{Q_{1}+Q_{2}}{4\pi(2R)^{2}} = \frac{Q_{1}+Q_{2}+Q_{3}}{4\pi(3R)^{2}}$
$\implies Q_{1} = \frac{Q_{1}+Q_{2}}{4} = \frac{Q_{1}+Q_{2}+Q_{3}}{9}$
$\implies Q_{2} = 3Q_{1}$ and $Q_{3} = 5Q_{1}$
$Q_{1} : Q_{2} : Q_{3} = 1 : 3 : 5$
$K\frac{Q^{2}}{4L^{2}} + K\frac{qQ}{L^{2}} = 0$
$q = -\frac{Q}{4}$
$qE = mg \implies E = \frac{mg}{q}$
$q = ne \implies n = \frac{3 \times 10^{-7}}{1.6 \times 10^{-19}}$
$\tau_{max} = P \times E = (2 \times 10^{-6})(0.03)(2 \times 10^{5}) = 12 \times 10^{-3} \text{ Nm}$
$E = 9 \times 10^{9} \frac{5 \times 10^{-9}}{(0.8)^{2}} = 7 \times 10^{4} \text{ N/C}$
$\tan \theta = \frac{qE}{mg}$
$\tan \theta = \frac{q}{mg}\left(\frac{\sigma}{2\epsilon_{0}}\right)$
$\implies \sigma \propto \tan \theta$
Dielectric constant is always $\ge 1$.
Charge flows only in the case of a potential difference.
Net charge $= 2q - q - q = 0$
Hence potential is zero, but the electric field is not zero.
$V = k\frac{q}{10} \implies kq = 10 \text{ V}$
$V^{\prime} = k\frac{q}{15} = \frac{10}{15}V = \frac{2}{3}V$
Potential at C:
$V_{c} = \frac{kq}{d/2} - \frac{kq}{d/2} = 0$
Detailed solution not required.
Detailed solution not required.
$C^{\prime} = n^{\frac{1}{3}}C$
$C^{\prime} = (8)^{\frac{1}{3}}C = 2C$
By the definition of dielectric strength.
If the battery is not connected, the charge will remain constant.
$C = \frac{\epsilon_{0}A}{d} = 10$
$C^{\prime} = \frac{\epsilon_{0}A}{d - t + \frac{t}{K}} = \frac{\epsilon_{0}A}{d - \frac{d}{2} + \frac{d}{2 \times 2}}$
$= \frac{\epsilon_{0}A}{\frac{3d}{4}} = \frac{4}{3} \times 10 = 13.33~\mu\text{F}$
$Q_{smaller}^{\prime} = \frac{R_{1}}{R_{1}+R_{2}}Q$
$= \frac{5}{15} \times 30 = 10~\mu\text{C}$
$W = \Delta U = U_{f} - U_{i}$
$= \frac{1}{2}\frac{q^{2}}{C/3} - \frac{1}{2}\frac{q^{2}}{C} = \frac{q^{2}}{C}$
$= \frac{C^{2}V_{0}^{2}}{C} = CV_{0}^{2} = \frac{\epsilon_{0}AV_{0}^{2}}{d}$
$C_{eq} = 2C = \frac{2\epsilon_{0}A}{d}$
$Q_{1} = Q_{2} + Q_{3}$ and $V_{2} = V_{3}$
$U_{before} = \frac{1}{2}\frac{q^{2}}{C_{1}}$
$U_{after} = \frac{1}{2}\frac{q^{2}}{C_{1}+C_{2}}$
$\frac{U_{before}}{U_{after}} = \frac{C_{1}+C_{2}}{C_{1}}$
$C_{1} = \frac{\epsilon_{0}A/2}{t/k_{1}} \implies C_{1} = \frac{\epsilon_{0}Ak_{1}}{2t}$ (Note: Depending on configuration, it could be series or parallel. From the given relation:)
$C_{1} = \frac{\epsilon_{0}A/2}{t/k_{1}} = \frac{\epsilon_{0}A k_{1}}{2t}$, $C_{2} = \frac{\epsilon_{0}A k_{2}}{2t}$
$C_{eq} = C_{1} + C_{2} = \frac{\epsilon_{0}A}{2t}[k_{1} + k_{2}]$
$i = \frac{2.5}{1 + 1 + 0.5} = 1\text{ A}$
$V = iR = 1 \times 2 = 2\text{ V}$
$q = CV = 5 \times 2 = 10~\mu\text{C}$
$C_{eq} = 1 + 1 = 2~\mu\text{F}$
Work done to bring charge $q = U_{2} - U_{1}$
$V = \frac{W}{q} = \frac{U_{2} - U_{1}}{q}$
$\vec{F}_{12} = K\frac{q_{1}q_{2}}{|\vec{r}_{21}|^{3}}\vec{r}_{21}$
$= 9 \times 10^{9} \frac{(-2 \times 10^{-6})(8 \times 10^{-6})}{(\sqrt{3})^{3}}(-\hat{i} + \hat{j} + \hat{k})$
$= \frac{16 \times 9 \times 10^{-3}}{3\sqrt{3}}(-\hat{i} + \hat{j} + \hat{k}) = 16\sqrt{3} \times 10^{-3}(-\hat{i} + \hat{j} + \hat{k})$
$\frac{q_{max}}{20 \times 10^{-4} \epsilon_{0}} = 3 \times 10^{6}$
$q_{max} = 60 \times 10^{2} \times 8.85 \times 10^{-12} = 529.2 \times 10^{-10} \approx 53 \times 10^{-9} \text{ C} \approx 53\text{ nC}$
Corresponding $E = 3 \times 10^{6} \times 2 \times 10^{-3} = 6000\text{ V}$
$\Delta V = -\int_{r_{1}}^{r_{2}} \vec{E} \cdot d\vec{r} = -\int_{10}^{20} \frac{100}{x^{2}} dx = -5$
So, difference $= 5\text{ V}$.
$\frac{q^{2}}{2r} - \frac{qQ}{r} - \frac{Qq}{r} = 0$
$\frac{q^{2}}{2} = 2qQ$
$\frac{q}{2} = 2Q \implies \frac{Q}{q} = \frac{1}{4}$
Detailed solution not required.
Detailed solution not required.
Detailed solution not required.
$q = CV = 50 \times 10^{-12} \times 10^{3} = 50 \times 10^{-9}\text{ C}$
$\frac{q}{A\epsilon_{0}K} \le 10^{7} \times \frac{10}{100}$
Putting the values, $A \ge 1.9 \times 10^{-3} \text{ m}^{2}$
Detailed solution not required.
Detailed solution not required.
Reverse checking:
$\vec{E} = -\left(\frac{\partial V}{\partial x}\hat{i} + \frac{\partial V}{\partial y}\hat{j}\right)$
Only option (2) gives the proper field.
Detailed solution not required.