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Three uniformly charged wires with linear charge density $\lambda$ are placed along $x,y$ and $z$. What is the flux of the electric field through Gaussian surface given by $$x^2 + y^2 + z^2 = 1;\, x>0;\, y>0;\, z>0\quad?$$

My Attempt:

The given equation represents a sphere with its center at the origin and radius equal to 1.

Since the given restrictions are $x>0;\, y>0;\, z>0$, I thought calculating the flux through the entire sphere and then dividing by 4 would give us the answer. I calculated the flux as,

$$\frac{3\lambda}{2\epsilon_0}$$

But the answer given is,

$$\frac{3\lambda}{4\epsilon_0}$$

where $\epsilon_0$ is the permittivity of free space. Any help would be appreciated.

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  • $\begingroup$ doesn't a sphere have 8 "quad"rants, or +++ out of 3-bits? $\endgroup$ – JEB Mar 22 at 22:07
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Since there are 8 quadrants in a sphere you'll get answer $$\frac{3\lambda}{4\epsilon_0}$$ I think you're assuming it to have 4 quadrants,Thus you are ending up with $$\frac{3\lambda}{2\epsilon_0}$$

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