midterm1_sol

# midterm1_sol - Name: Student ID number: SOLUTIONS 1 2 3 4...

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Name: SOLUTIONS Student ID number: 1 2 3 4 P A [1.] (30 pts) Spherical Capacitor. (the outer shell) extends from d < shell is given charge - Q R o a b c d e +Q -Q (a) Draw a (clear) diagram indicating the location of surface charges (or if you don’t like to draw, state it). Indicate the total charge on each surface. (b) What is the electric ﬁeld everywhere in space? Plot E versus r . (c) What is the capacitance of this system (use the innermost sphere and the outermost shell as the “plates” of the capacitor)?

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Solution (a) I’ll be lazy and just state where the charges are: On the inner sphere, the + Q charge is spread evenly on the surface. The middle spherical shell polarizes – negative charge gathers on the inner surface of the shell (value - Q ) and positive charge is exposed on the outer surface (value + Q ). On the outer conducting shell, the charge - Q gathers on the inner surface, and there is no charge on the outer surface. The charges arrange themselves in this way to make the electric ﬁeld zero inside each conductor. You can use Gauss’ law to show this rigorously – you want to make E = 0 in side of each conductor, and with spherical symmetry, this means the charge enclosed by a spherical Gaussian surface which sits inside of any of the conductors must be zero. (b) We know that the electric ﬁeld is zero inside of each conductor (the surface charges take care of this). So we need to worry about the regions in between the conductors and the region outside of the outer spherical shell. Note that all of these regions are vacuum regions – we know that the ﬁeld is going to look like that of a point charge. All we need to know to write down the ﬁeld in these regions is the total charge enclosed by an imaginary Gaussian surface in these regions. We start with the region r > e , outside of the outer shell. The total charge enclosed by a spherical Gaussian surface in this region is zero – there is + Q on the inner sphere and - Q on the outer (and no net charge on the middle). So the ﬁeld is zero for r > e . For c < r < d , the total charge enclosed is + Q , as there is no net charge on the middle shell. Therefore: E r = Q 4 πe o r 2 c < r < d For a < r < b , the total charge enclosed is again + Q so: E r = Q 4 o r 2 a < r < b So now for the plot: (c) To ﬁnd the capacitance, we need to ﬁnd the potential difference between our plates, given that one has charge + Q and the second has charge - Q (which is what we found the electric ﬁeld for above). We integrate from the inner edge of the outer shell to the outer edge of the inner sphere: V ( a ) - V ( d ) = - Z a d E r dr = - Z c d Q 4 o r 2 dr - Z a b Q 4 o r 2 dr 2
The simpliﬁcation above comes from the fact that the electric ﬁeld is zero inside of the middle conducting shell. The answer is: Δ V = V ( a ) - V ( d ) = Q 4 πe o ± 1 a + 1 c - 1 b - 1 d ² We then use Q = C Δ V to get the capacitance: C = 4 o ± 1 a + 1 c - 1 b - 1 d ² - 1 3

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[2.] (30 pts) Point charge and grounded shell. A point charge of value + q is placed inside a grounded conducting shell of inner radius a and outer radius c . As shown in the ﬁgure
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## This test prep was uploaded on 04/18/2008 for the course PHYS 110a taught by Professor Miao during the Winter '08 term at UCLA.

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midterm1_sol - Name: Student ID number: SOLUTIONS 1 2 3 4...

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