An infinity long uniform line charge distribution of charge per unit length `lambda` lies parallel to the y-axis in the `y-z` plane at `z=sqrt3/2` a (see figure). If the magnitude of the flux of the electric field through the rectangular surface ABCD lying in the `x-y` plane with its centre at the origin is `(lambdaL)/(n epsilon_0)` (`epsilon_0=` permittivity of free space), then the value of n is

Answer

6

Explanation

Here, `E = (lambda)/(2pi in_(0) r) = (lambda)/(2pi in_(0) sqrt(x^(2) + (3a^(2))/(4)))`
`d phi = E sin theta . A cos 0 = E sin theta . A`
`= (lambda)/(2pi in_(0) sqrt(x^(2) + (3a^(2))/(4))) xx (sqrt(3) a//2)/(sqrt(x^(2) + (3a^(2))/(4))) xx Ldx`
`= ((lambda sqrt(3) a//2) Ldx)/(2pi in_(0) (x^(2) + (3a^(2))/(4)))`
`(lambda)/(2pi in_(0)) [((sqrt(3) a)/(2))/(x^(2) + (3a^(2))/(4))] Ldx`

`= (lambda)/(2pi in_(0)) [(sqrt(3) a//2)/((x^(2) + (3a^(2))/(4))/((3a^(2))/(4)) xx (3a^(2))/(4))] Ldx`
or `d phi = (lambda)/(2pi in_(0)) [(sqrt(3) a//2)/([(x^(2))/(3a^(2)//4) + 1] (sqrt(3) a^(2))/(2))] Ldx`
`= (lambda)/(2pi in_(0)) [(sqrt(3) a//2)/([(x^(2))/(3a^(2)//4) + 1] [(sqrt(3) a)/(2)]^(2)]] Ldx`
`phi = int d phi = int_(-a//2)^(a//2) (lambda)/(2pi in_(0)) (1)/([(x^(2))/(3a^(2)//4) + 1] [(sqrt(3) a)/(2)]) Ldx` ...(i)
Let `(x)/(sqrt(3) a//2) = y`
or `x = (sqrt(3) a)/(2) y`
or `dx = (sqrt(3) a)/(2) dy`
or `(dx)/(sqrt(3) a//2) = dy`
From (i), `phi = (lambda L)/(2pi in_(0)) int_(-1//sqrt(3))^(+1//sqrt(3)) (dy)/(y^(2) + 1)`
`[[x= (a)/(2)" corresponds to "y = - (1)/(sqrt(3))],[x = + (a)/(2)" corresponds to "y = (1)/(sqrt(3))],[f rom x = (sqrt(3)a)/(2) y]]`
`:. phi = (lambda L)/(2pi in_(0)) xx2 tan ^(-1) ((1)/(sqrt(3)))`
`= (lambda L)/(2pi in_(0)) xx2xx (pi)/(6) = (lambda L)/(6 in_(0))`
Comparing with `phi = (lambda L)/(n in_(0))`, we get `n = 6`