Physics Electrostatic Potential and Capacitance

(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

Before closing the switch

After closing the switch

So charge on C₁  remains same, on C₂ increases and on C₃ and C₄  decreases.

Please find the answer the below

Please find the answer below

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.

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

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

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

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

$=50\%$  

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

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

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

$\Rightarrow Workdone=\Delta K$

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

This chapter covers the concepts of potential, capacitors, and potential energy. It is considered as one of the easy chapter of the class 12 Physics.

It is the farad (F). The name came from Michael Faraday. SI Unit of Capacitance means the ability of a system to store an electric charge. One farad is the capacitance of a device that needs one coulomb of charge to provide one volt potential difference across it. Mathematically, it is represented by - 1 Farad = 1 Coulomb/Volt.