Question

Question in attachment PYQ
#JEE mains !​

Answer 1

Electric field as a function of position (r) is given as :

$\vec{E} = Ar {}^{ \frac{3}{2} }$

From Guass law,

$\displaystyle \: \oint \vec{E}. \vec{ds} = \dfrac{q}{ \epsilon_o} \\ \\ \implies \epsilon_o( Ar {}^{ \frac{3}{2} } ) \oint ds = q \\ \\ \implies q = \epsilon_o( Ar {}^{ \frac{3}{2} } )(4\pi {r}^{2} ) \\ \\ \implies q = 4\pi\epsilon_oAr {}^{ \frac{5}{2} }$

At 'R' distance from origin (r = R),

$\implies \boxed{ \boxed{q = 4\pi\epsilon_oA {R}^{ \frac{5}{2} } }}$

Option (b) is correct

Answer 2

Explanation:

topic :

• electric field

given :

• An electric field in a certain region is radially outwards from origin and has a magnitude equal to A³/2, where r is distance from the origin. The charge contained in a sphere of radius R, centered around origin is

to find :

• The charge contained in a sphere of radius R, centered around origin is

solution :

• E.ds = q/e

• → €₁ (Ar³) f ds = q

• q = €₁ (Ar ³) (4πr²2)

• q= 4π€, Ar 5/2

• sphere of radius R = (r = R)

• q = 4T€, AR 5/2

• option b is correct