Question

Derive :

$\bf{\dfrac{1}{4\pi \varepsilon_{0}} \dfrac{Q}{r^{2}}}$

Answer with a explanation :)

​

Answer 1

Derivation

Electric field due to +ve charge :- away from the charge

Electric field due to -ve charge :- toward from the charge

$\rm \: q_o(very \: very \: small) \: is \: test \: charge$

$\rm \: limit \: q_o \to \: 0$

so that q₀ do not creat any electric field of its own

intensity of electric field due to a point charge + q at distance r

To Find electric field we place one test charge

$\rm \:\overrightarrow{E}= \dfrac{force \: on \: test \: charge}{test \: charge}$

where

$\rm\overrightarrow{E} = intensity \: \: field$

For +q

$\rm\overrightarrow{E} = \frac{f}{q_o}$

Putting the value

$\rm\overrightarrow{E} = \dfrac{kqq_o}{r {}^{2} q_o}$

$\rm\overrightarrow{E} = \dfrac{kq \cancel q_o}{r {}^{2} \cancel q_o}$

$\rm\overrightarrow{E} = \dfrac{kq}{r {}^{2} }$

Away from point

Electric field intensity due to a point charge at distance ' r ' is

$\boxed{ \rm \: E = \dfrac{Kq}{ {r}^{2} } }$

SI unit = N / C

when

$K = \dfrac{1}{4\pi\varepsilon_{0}}$

$\boxed{ \rm \: E = \dfrac{q}{4\pi\varepsilon_{0} {r}^{2} } }$

Note

=> this formula valid for point charge not valid on continuous/ extended charge system

=> The electric field due to a point charge of its own location is not defined