A disk of radius $\frac{a}{4}$ having a uniformly distributed charge 6 C is placed in the x-y plane with its centre at $(- \frac{a}{2}, 0, 0)$ . A rod of length ‘a’ carrying a uniformly distributed charge 8C is placed on the x-axis from $x = \frac{a}{4}$ to $x = \frac{5 a}{4}$ . Two point charges -7C and 3C are placed at $(\frac{a}{4}, - \frac{a}{4}, 0) and (- \frac{3 a}{4}, \frac{3 a}{4}, 0)$ respectively.
Consider a cubical surface formed by six surfaces $x = \pm \frac{a}{2}, y = \pm \frac{a}{2}, z = \pm \frac{a}{2}$ . The electric flux through this cubical surface is

Moderate

Solution

From Gauss' Law, we have flux equal to
$ϕ = \frac{{ΣQ}_{enc}}{ε_{0}} = \frac{(\frac{8 C}{4}) - 7 C + (\frac{6 C}{2})}{ε_{0}} = - \frac{2 C}{ε_{0}}$

OR
The disc is placed with its center at (-a/2, 0, 0). Hence we can say that half of the discs are enclosed within the cube. Therefore, the charge on the disc enclosed within the cube is 6C/2= 3C.
The length of the rod is ’a’ with total charge 8C. The surface of the cube lies at a distance of a/2. Therefore the section of the rod enclosed within the cube is a/2-a/4=a/4.Therefore the total charge of the rod enclosed within the cubical surface is 2C. It is to be noted that the charge 3C lies outside the cubical surface. The charge -7C lies within the cubical surface i.e. at (a/4, -a/4, 0).
From the above information, the total charge (Q)enclosed within the surface is,
Q=3C+2C+−7CQ=−2C
$ϕ = - \frac{2 C}{ε_{0}}$

Get Instant Solutions

When in doubt download our app. Now available Google Play Store- Doubts App

Recieve an sms with download link

SCAN ME