Answer to part 1 was 0 Examine the 3-D conductor in the figure, which has a cavi

Question

Answer to part 1 was 0

Examine the 3-D conductor in the figure, which has a cavity in the center. The conductor has an excess charge 5.1 mu C on it. What is the electric flux (in N middot m^2/C) through the Gaussian surface S_1 shown in the figure? Now put a point of charge 28 mu C inside the cavity of the conductor. What is the flux (in N middot m^2/C) through the Gaussian surface S1? Now consider the Gaussian surface S_2. With the charge still inside the cavity, what is the flux (in Nm^2/C) through this surface?

Explanation / Answer

(a) According to Gauss's law electric flux through a surface = Q/epsilon

Where Q is the charge enclosed by that surface

Now in part a charge is given to the conductor so all charge will move to its surface and no charge will remain inside the surface S1 so electric flux will be zero through S1

Flux =0

(b) Now 28 uC charge is placed inside the cavity so charge enclosed by S1 will be equal to 28uC

Flux = (28 x 10-6)/(8.85 x 10-12)

Flux = 3.16 x 106 Nm2/C

(c) Now for S2 total charge inside this surface is 28uC and 5.1uC

Q = (28 x 10-6)+(5.1 x 10-6)

Q = 33.1 x 10-6 C

Flux = (33.1 x 10-6)/(8.85 x 10-12)

Flux = 3.74 x 106 Nm2/C

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