Question

Electric field intensity due to point charge q at a distance r varies as

Answer 1

Answer:

Point charge q is placed at a specified position named as Origin on the coordinate axis (0,0)

As per a Spherical Gaussian Surface around the point charge at a distance of r , we can say that:

$\displaystyle \int \:E \: .ds = \frac{q}{ \epsilon_{0}}$

Since the electric field vector and area vector for elemental area is zero degree (0°), we can say :

$= > \displaystyle \int \:E \: ds \: \times \{ \cos( 0 \degree) \} = \frac{q}{ \epsilon_{0}}$

$= > \displaystyle E\int ds \: \times \{ 1 \} = \frac{q}{ \epsilon_{0}}$

$= > E(4\pi {r}^{2} ) = \dfrac{q}{ \epsilon_{0}}$

$= > E= \dfrac{q}{4\pi \epsilon_{0} {r}^{2} }$

$= > E= \dfrac{q}{4\pi \epsilon_{0} } \bigg \{ \dfrac{1}{ {r}^{2} } \bigg \}$

Considering charge to be constant , we can say that:

$= > E \: \propto \: \dfrac{1}{ {r}^{2} }$

So, final answer :

$\boxed{ \huge{ \red{ E \: \propto \: \dfrac{1}{ {r}^{2} } }}}$