JEE Main 27-Aug-2021 Shift 1 Solved Paper

A circular disc is placed in xy-plane with its centre at the origin as shown below   Consider an elemental ring of thickness dr and radius r. Now, the area of the elemental ring can be given by
dA=2π‌rdr
The charge stored in this elemental ring,
dq=σ‌dA
Now, the electric field at the point on Z-axis at a distance of Z from origin can be given by
dE=

kdqZ

(r2+Z2) 3 2

Substituting the value of dq and dA in above equation, we get
dE=

kZ‌σ(2π‌rdr)

(r2+Z2) 3 2

=

σZ

2ε0

(

rdr

(r2+Z2) 3 2

)
Calculating the total electric field by integrating the above expression from r = 0 to r = R, we get
E=σZ2ε0‌∫0R

rdr

(r2+Z2) 3 2

Put r2+Z2=u2
⇒2‌rdr=2‌udu
⇒rdr=ud
For lower limit, r = 0
⇒ u = Z
Upper limit, r = R
⇒u=√R2+Z2
E=

σZ

2ε0

‌∫Z√R2+Z2

udu

u3

E=

σZ

2ε0

‌∫Z√R2+Z2

du

u2

=

σZ

2ε0

‌∫Z√R2+Z2u−2du
=

σZ

2ε0

[

u−2+1

(−2+1)

]Z√Z2+R2
=

σZ

2ε0

[−

1

u

]Z√Z2+R2
=

σZ

2ε0

[−

1

√(Z2+R2)

+

1

Z

]
=

σZ

2ε0

[

1

Z

−

1

√(Z2+R2)

]
=

σ

2ε0

[1−

Z

√(Z2+R2)

]
Thus, the electric field at the point on Z-axis at a distance of Z from origin is

σ

2ε0

[1−

Z

√(Z2+R2)

]