Q.

A uniformly charged disc of radius R having surface charge density σ is placed in the xy-plane with its centre at the origin. Find the electric field intensity along the Z-axis at a distance Z from origin

see full answer

Start JEE / NEET / Foundation preparation at rupees 99/day !!

21% of IItians & 23% of AIIMS delhi doctors are from Sri Chaitanya institute !!
An Intiative by Sri Chaitanya

a

E=2ε0σ1Z2+R2+Z

b

E=σ2ε01ZZ2+R2

c

E=σ2ε01+ZZ2+R2

d

E=σ2ε01Z2+R2+1Z2

answer is A.

(Unlock A.I Detailed Solution for FREE)

Ready to Test Your Skills?

Check your Performance Today with our Free Mock Test used by Toppers!

Take Free Test

Detailed Solution

A circular disc is placed in xy-plane with its centre at the origin as shown below

Question Image

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=kdqZr2+Z232

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

dE=kZσ(2πrdr)r2+Z23/2=σZ2ε0rdrr2+Z23/2

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

       E=σZ2ε00Rrdrr2+Z23/2 Put r2+Z2=u22rdr=2udurdr=udu

For lower limit, r = 0   ⇒u = Z

Upper limit,  r=Ru=R2+Z2E=σZ2ε0ZR2+Z2uduu3E=σZ2ε0ZR2+Z2duu2=σZ2ε0ZR2+Z2u2du=σZ2ε0u2+1(2+1)ZZ2+R2=σZ2ε01uZZ2+R2=σZ2ε01Z2+R2+1Z=σZ2ε01Z1Z2+R2=σ2ε01ZZ2+R2

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

σ2ε01ZZ2+R2

Watch 3-min video & get full concept clarity

tricks from toppers of Infinity Learn

score_test_img

Get Expert Academic Guidance – Connect with a Counselor Today!

whats app icon