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The closed Gaussian surface shown at right consists of a hemispherical surface a

ID: 2168383 • Letter: T

Question

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. What is the flux through the entire closed surface? Explain. Let Phi L represent the flux through the flat left-hand portion of the surface. Write an expression in terms of Phi L for the flux through the cursed portion of the surface, Phi c. Suppose that the curved portion of the Gaussian surface in part a is replaced by the larger cursed surface as shown. The flat left-hand portion of the surface is unchanged. Does the value of Phi L change? Explain. How does the flux through the new curved portion of the surface compare to the flux through the original curved portion of the surface? Explain. Suppose that the curved portion of the Gaussian surface is replaced by the larger curved surface that encloses the charge as shown. The flat left-hand portion of the surface is still unchanged. Docs the value of Phi L change? Explain. How does the flux through the new curved portion of the surface compare to the flux through the original curved portion of the surface? Explain. Use Gauss' law to write an expression in terms of Phi L and q for the flux through the curved portion of the surface.

Explanation / Answer

i) flux = 0
because from gauss law
flux through a closed surface = charge enclosed by the surface/

here , charge enclosed =0

hence flux = 0

ii) fom part (i)

we know that,

total flux = 0

L + c = 0

c = -L

(iii) L does not change because the left surface has not changed.

flux through this new surface , c_new must satisfy

c_new + L =0 ( from gauss law)

c_new =- L

hence, it remains the same

(iv)

L will remain same, because the left hand portion of surface is unchanged with respect to q

however,

from gauss law

c + L =q/ (beacuse the surface now encloses q)

c = q/ - L

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