Question

[$σ$ is the surface charge density on the disc]

Open in App

Solution

Verified by Toppr

Correct option is A)

We can assume the charge distribution as a collection of concentric rings of charge. Consider one such ring of radius $r$ and charge $dq$.

Area element $dA=(2πr)dr=$ area of ring

$dq=σdA=(2πσr)dr=$ charge of that ring

Due to symmetry, there is no vertical component of electric field at point P. There is only the horizontal component so, using the result from a ring of charge;

$dE_{x}=(x_{2}+r_{2})_{3/2}k(dq)x $

$=(x_{2}+r_{2})_{3/2}k(2πσr)drx $

$E_{x}=∫_{0}dE_{x}=∫_{0}(x_{2}+r_{2})_{3/2}(2πσrkdr)x $

put $x_{2}+r_{2}=t_{2}$

Differentiating both sides,

$2rdr=2tdt$

At $r=0;t=x$

At $r=a;t=x_{2}+a_{2} $

So, $E_{x}=2ε_{0}σ ∫_{x}t_{3}t dt$

$=2ε_{0}σ [t−1 ]_{x}$

$=2ε_{0}σ [x1 −x_{2}+a_{2} 1 ]$

$=2ε_{0}σ ⎣⎢⎢⎢⎡ 1−x_{2}a_{2} +1 1 ⎦⎥⎥⎥⎤ $ is very small compared to 1.

So, $E_{x}=2ε_{0}σ $.

Was this answer helpful?

0

0