Gauss Law The closed Gaussian surface shown at right consists of a hemispherical
ID: 2075432 • Letter: G
Question
Gauss Law
The closed Gaussian surface shown at right consists of a hemispherical surface and a flat plane. A point charge +q is outside the surface, and no charge is enclosed by the surface. a. What is he flux through the entire closed surface? Explain. Let Phi_F represent the flux through the flat (left-hand) portion of the surface. Write an expression in terms of Phi_F for the flux through the curved portion of the surface, Phi_C. b. Suppose that the curved portion of the Gaussian surface in part a is replaced by a larger curved surface as shown. The flat portion of the surface is unchanged. i. Does the value of Phi_F change? Explain. ii. Does the value of Phi_C change? Explain. c. Suppose that the curved portion of the Gaussian surface is replaced by a larger curved surface that encloses the charge as shown. The flat portion of the surface is still unchanged. i. Does the value of Phi_F change? Explain. ii. Does the value of Phi_C change? Explain. iii. Use Gauss' law to write an expression for Phi_C in terms of Phi_F and q.Explanation / Answer
Flux through the entire closed volume will be zero as according to Gauss law, charge should be enclosed inside the Gaussian Surface.
The flux lines entering inside the Gaussian Surface exists the Surface from some other point, so net Flux remains zero.
Therefore negative {PHIc} is equal to {PHIF}
B.
No the values of both donot change, as charge is still not insode the volume of Gaussian Surface, and net Flux is again zero.
C.
Yes the values change for curved surface. as charge is now enclosed inside the Gaussian Surface
But for flat surface, there will be no change in flux through it. As charge is on same side as it was earlier, and same flux lines woupd pass through flat surface.
Net flux is given by q/(epsilon0)
And through flat surface is {PHIF}, so through curved surface will be : q/(epsilon0)- {PHIF}
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