Electrostatics

$U_{\text{initial}}=\frac{k\left(4q\right)\left(q\right)}{(d/2)}+\frac{k\left(q\right)(-q)}{(d/2)}$

$\begin{array}{rr}& \Rightarrow \mathrm{}\frac{6k{q}^2}{d}\\ & \Rightarrow \mathrm{}{U}_{\text{final}}=\frac{4\left(4q\right)\left(q\right)}{\left(\frac{3d}{2}\right)}+\frac{k\left(q\right)(-q)}{(d/2)}\\ & \Rightarrow \mathrm{}\frac{2}{3}\frac{k{q}^2}{d}\\ & \Rightarrow \mathrm{}\mathrm{\Delta }U=\left(\frac{2}{3}-6\right)\frac{k{q}^2}{d}\Rightarrow \frac{-16}{3}\frac{{\mathrm{k}\mathrm{q}}^2}{d}\end{array}$

${\sigma }_1={\sigma }_2$

$\begin{array}{rr}& \frac{{\mathrm{Q}}_1}{4\pi {\mathrm{R}}^2}=\frac{{\mathrm{Q}}_2+{\mathrm{Q}}_1}{4\pi \left(16{\mathrm{R}}^2\right)}\\ & \Rightarrow \mathrm{}{\mathrm{Q}}_1+{\mathrm{Q}}_2=16{\mathrm{Q}}_1\\ & \Rightarrow \mathrm{}{1}_1={\mathrm{Q}}_2\\ & \mathrm{V}\left(\mathrm{R}\right)-\mathrm{V}\left(4\mathrm{R}\right)\\ & =\left(\frac{{\mathrm{K}\mathrm{Q}}_1}{\mathrm{R}}+\frac{{\mathrm{K}\mathrm{Q}}_2}{4\mathrm{R}}\right)-\left(\frac{{\mathrm{K}\mathrm{Q}}_1}{4\mathrm{R}}+\frac{{\mathrm{K}\mathrm{Q}}_2}{4\mathrm{R}}\right)\\ & =\frac{3\mathrm{K}\mathrm{Q}{\mathrm{Q}}_1}{4\mathrm{R}}=\frac{3{\mathrm{Q}}_1}{16\pi \in \mathrm{R}}\end{array}$

When the two conducting spheres are connected by a wire, they will reach the same electric potential, V.
The total charge Q_total = 12µC + (-3µC) = 9µC. This total charge will redistribute.
Let the final charges be q? and q? + q? = 9µC.
The potential of a sphere is V = kq/r.
V? = V?
k q? /R? = k q? /R? ⇒ q? /R? = q? /R?
q? / (2R/3) = q? / (R/3) ⇒ q? /2 = q? ⇒ q? = 2q?
Substitute this into the charge conservation equation:
2q? + q? = 9µC ⇒ 3q? = 9µC ⇒ q? = 3µC.
Then, q? = 2 * 3µC = 6µC.
The final charges are 6µC and 3µC.

Heat loss; ΔH = U? - U? = 1/2 (C? / (C? +C? ) (V? -V? )²
= 1/2 * (5*2.5)/ (5+2.5) (220-0)²µJ
= 5/ (2*3) * 22*22*100*10? J
= 5*11*22/3 * 10? J = 1210/3 * 10? J = 1210/3 * 10? ³ J * 4 * 10? ²
According to questions
x/100 = 4*10? ²
So, x=4
Note: But given answer by JEE Main x=36

Before inserting slab
C_i = ε? A/d
After inserting dielectric slab
C_i = ε? lw/d
C_f = C? + C?
C_f = (Kε? A? /d) + (ε? A? /d)
C_f = 2C_i ⇒ (Kε? wx/d) + (ε? w (l-x)/d) = 2ε? lw/d
4x + l - x = 2l
x = l/3

Potential of centre, = V =
Vc = K (Σq)/R
Vc = K (0)/R = 0
Electric field at centre E_B = E_B = ΣE
Let E be electric field produced by each

charge at the centre, then resultant electric field will be
Ec = 0, since equal electric field vectors are acting at equal angle so their resultant is equal to zero.