JEE Advanced Model Paper 15 with solutions for online practice

For the considered region given charge distribution will act as a diole system. Therefore electric field anywhere is the region is directed towards +ve x-axis.Due to symmetry we can conclude that electric potential anywhere in y-z plane will be zero
So the work done from bringing a positive test change from (0,

a

√2

,

a

√2

) to (a,

a

√2

,

a

√2

) is zero
Electric flux crossing

Φ =

b

∫

r=a

E (r) (2πrdr) = (kP) (2π)

∫

b

dr

r2

a
=

1

4πε0

(2qd) (2π) (

−1

r

)ab = qd/∊0 (

1

a

−

1

b

)