Question

Application of guass law​

Answer 1

$\huge{ \underline{\underline{ \fcolorbox{white}{pink}{\sf{Answer :-}}}}}$

Applications of Gauss Law

1. In the case of a charged ring of radius R on its axis at a distance x from the centre of the ring. E = \frac{1}{4\pi {{\in }_{0}}}\frac{qx}{{{\left( {{R}^{2}}+{{x}^{2}} \right)}^{3/2}}}

4π∈

0

1

(R

2

+x

2

)

3/2

qx

. At the centre, x = 0 and E = 0.

2. In case of an infinite line of charge, at a distance ‘r’. E = (1/4 × πrε0) (2π/r) = λ/2πrε0. Where λ is the linear charge density.

3. The intensity of the electric field near a plane sheet of charge is E = σ/2ε0K where σ = surface charge density.

4. The intensity of the electric field near a plane charged conductor E = σ/Kε0 in a medium of dielectric constant K. If the dielectric medium is air, then Eair = σ/ε0.

5. The field between two parallel plates of a condenser is E = σ/ε0, where σ is the surface charge density.

Answer 2

$\huge\mathcal{\boxed{\underline{\color{olive} Good\: Morning}}}$

$\huge\mathcal{\underline{\color{aqua}QUESTION}}$

⚽ Application of guass law .

$\huge{\orange{\boxed{\fcolorbox{lime}{cadetblue}{\pink{ANSWER}}}}}$

✨ Gauss Law states that the total electric flux out of a closed surface is equal to the charge enclosed divided by the permittivity. The electric flux in an area is defined as the electric field multiplied by the area of the surface projected in a plane and perpendicular to the field.

$\pink\longmapsto~\bf\green{\oint~\vec{E}~.~\vec{ds}~=~\dfrac{q}{\epsilon_o}~}$

APPLICATION ;-

1. In the case of a charged ring of radius R on its axis at a distance x from the centre of the ring.

• $\bf\red{E~=~\dfrac{1}{4~\pi~\epsilon_o}.\dfrac{qx}{(R^2~+~x^2)^{\tiny\dfrac{3}{2}}~}}$ At the centre, x = 0 and E = 0.

2. In case of an infinite line of charge, at a distance ‘r’.

✔️ E = (1/4 × πrε0) (2π/r) = λ/2πrε0.

Where

• λ is the linear charge density.

3. The intensity of the electric field near a plane sheet of charge is E = σ/2ε0K

where

• σ = surface charge density.

4. The intensity of the electric field near a plane charged conductor E = σ/Kε0 in a medium of dielectric constant K. If the dielectric medium is air, then

• $\bf\gray{E_{air}}$ = σ/ε0

5. The field between two parallel plates of a condenser is E = σ/ε0,

where

• σ is the surface charge density.