Electrostatic Potential and Capacitance

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. 

In JEE Main Physics, electric potential and capacitance chapter has a weightage of 3% to 6%. You can expect around 1 or 2 questions from this chapter which carries 4 to 8 marks.

In NEET Physics exam, the chapter electric potential and capacitance carries a weightage of 2% to 5%, which means you can expect one question out of 45 questions.

The ability to do work on a charge is called the electric potential and the ability to store charge is termed as the capacitance.

After switch 'S' is closed

$Q_1+Q_2=C_{1}V$ -    (1)

              Using KVL

              $\frac{Q_1}{C_1}-\frac{Q_2}{C_2}=0$  

              $Q_1=Q_2\frac{C_1}{C_2}$                       - (2)

              from (1) & (2)

              $\Rightarrow Q_2\left[\frac{C_1+C_2}{C_2}\right]=C_{1}V\Rightarrow Q_2=\left(\frac{C_1{C}_2}{C_1+C_2}\right)V$

E = ½CV²
= ½ (ε? A/d)Ed²
= ½ε? E²Ad

$\mathrm{V}=\frac{1}{4\pi {?}_0}\cdot \frac{\mathrm{Q}}{\mathrm{R}}$

$\therefore \mathrm{V}\propto \frac{1}{\mathrm{R}}$

$\therefore$  Potential is more on smaller sphere.

Potential difference across resistor at time t = V = 30/3 = 10V
Current, I = 10 / (5 * 10? ) = 2? A

1/C = 5/ (ε? * 100) + 5/ (10ε? * 100)
⇒ C = (ε? * 1000) / 55 = (8.85 * 10? ¹² * 1000) / 55 = 1.61 * 10? ¹? F = 161 pF