Please provide all the formula related to 1st and 2nd chapter of physics class 12 "Electric charges and fields" and Electrostatic potential and capacitance". Do not say that refer the study material or revision notes.

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Please find below the solution to the asked query:

Here i am providing some of the important formulae, but i strongly suggest you, please study from books. Because, in order to solve the problems, you have to know the interrelation between the topics. If you try to remember the formulae, you will not be able to apply for the problems. Some of the topics i am entering slide images, because, this is too lengthy question. Typing is too difficult. If you want to know as typed, Please ask topic by topic.

Electric Charge:

Coulumb's Law:

• Electric force between two point charges :      $\overrightarrow{F_{12}}=\frac{1}{4\pi {\epsilon }_{0}K}\frac{q_1{q}_2}{r_{12}^2}\widehat{r_{12}}=-\overrightarrow{F_{21}}$

Where,
where K = 1 for free space​, $\frac{1}{4\pi {\epsilon }_{0}K}=9\times {10}^9 N m^2 C^{-2}$

• Coulomb's law agrees with Newton's third law.
• Force on charge q₁ due to remaining charges q₂, q₃, q₄, ... q_n in the region is ; 

$\overrightarrow{F_1}=\frac{q_1}{4\pi {\epsilon }_0}\sum _{i=1}^n\frac{q_i}{r_{1i}^2}\widehat{r_{1i}}$
It is a vector sum of all the forces acting on point charge q₁.​

Electric Field:

• Electric flux through a plane surface area $∆S$ held in a uniform electric field​

$∆{\varphi }_E=\overrightarrow{E}.∆\overrightarrow{S}=E∆S \cos \theta$

• According to Gauss's law, electric flux through a closed surface S is equal to $\left(\frac{1}{{\epsilon }_0}\right)$ times charge enclosed.

${\varphi }_E=\oint \overrightarrow{E}.d\overrightarrow{S}=\frac{q_{en}}{{\epsilon }_0}$

• Electric field due to a long straight wire of uniform linear charge density λ is $E=\frac{\lambda }{2\pi {\epsilon }_{0}r}$
• Electric field due to an infinite plane sheet of uniform surface charge density σ is $E=\frac{\sigma }{2{\epsilon }_0}$
• Electric field due to two equally and oppositely charged parallel plates is 

$E=0 \left(for outside points\right); E=\frac{\sigma }{{\epsilon }_0}\left(for inside points\right)$

• Electric field due to a thin spherical shell of charge density ρ and radius R is

$$\begin{gathered} E=\frac{1}{4\pi {\epsilon }_0}\frac{q}{r^2} \left(for r>R\right) \\ E=\frac{1}{4\pi {\epsilon }_0}\frac{qr}{R^3} \left(for r<R\right) \\ E=\frac{1}{4\pi {\epsilon }_0}\frac{q}{R^2} \left(for r=R\right) \\ Where, \\ q=\frac{4}{3}\pi R^3\rho \end{gathered}$$

Electric Potential:



Potential of charge distributions:



Electrostatic Potential Energy:



Electric Dipole:



Dielectrics and Polarisation:



Capacitors and Capacitance:



Different types of Capacitors:



Combination of Capacitors:


 

 

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