Electrostatics

$C_1=1\mu F{V}_1=100V?U_1=\frac{1}{2}*1*{10}^{-6}*{\left(100\right)}^2=5*{10}^{-3}J$  

$C_2=1\mu F$ , common potential $V=\frac{C_1{V}_1}{C_1{C}_2}=\frac{100}{2}=50V$

Final electrostatic energy stored in both the capacitors.

$=\frac{1}{2}\left(C_1+C_2\right)V^2=\frac{1}{2}*2*{\left(50\right)}^2*{10}^{-6}$

$=2.5*{10}^{-3}J$

$Energylost=2.5*{10}^{-3}$

$\%lossofenergy=\frac{2.5*{10}^{-3}}{5*{10}^{-3}}*100$

$=50\%$

From work energy theorem

$W_{sp}+W_E=?K$

$-\frac{1}{2}k{x}_0^2+QE.x_0=0$

$?x_0=\frac{2QE}{k}$

(Explanation can be elaborate)

For a region of constant electric potential

(1) The electric field may be uniform

(2) The electric field may be zero

(3) There can be no charge inside the region

Direction of  is perpendicular to equipotential surface.

$C_1=1\mu F{V}_1=100V?U_1=\frac{1}{2}*1*{10}^{-6}*{\left(100\right)}^2=5*{10}^{-3}J$  

$C_2=1\mu F$ , common potential $V=\frac{C_1{V}_1}{C_1{C}_2}=\frac{100}{2}=50V$

Final electrostatic energy stored in both the capacitors.

$=\frac{1}{2}\left(C_1+C_2\right)V^2=\frac{1}{2}*2*{\left(50\right)}^2*{10}^{-6}$

$=2.5*{10}^{-3}J$

$Energylost=2.5*{10}^{-3}$

$\%lossofenergy=\frac{2.5*{10}^{-3}}{5*{10}^{-3}}*100$

$=50\%$  

$C_{eq}=2C=\frac{2A{\epsilon }_0}{d}$

$F=qEa=\frac{qE}{m}$

$\Rightarrow s=\frac{1}{2}a{t}^2=\frac{1}{2}\left(\frac{qE}{m}\right)t^2$

$\Rightarrow Workdone=\Delta K$

$\Rightarrow \Delta K=F.s=\frac{q^2{E}^2{t}^2}{2m}$

For a region of constant electric potential

(1) The electric field may be uniform

(2) The electric field may be zero

(3) There can be no charge inside the region

 Kindly go through the solution

C = 3A? ₀/d

With the help of conservation of volume, we can write

  $27*\frac{4}{3}\pi r^3=\frac{4}{3}\pi R^3\Rightarrow R=3r.......\left(1\right)$                            

With the help of conservation of charge, we can write

Q = 27 q.(2)

Potential energy of single drop = U₁ =   $\frac{q^2}{8\pi {\epsilon }_{0}r}$

Potential energy of bigger drop =   $U_2=\frac{Q^2}{8\pi {\epsilon }_{0}R}=\frac{27*27*q^2}{8\pi {\epsilon }_0\left(3r\right)}=243\left(\frac{q^2}{8\pi {\epsilon }_{0}r}\right)=243U_1$

$\Rightarrow \frac{U_2}{U_1}=243$