874 when an excess charge is placed on the surface of a soap bubble, its radius

Question

874 when an excess charge is placed on the surface of a soap bubble, its radius increases because of repulsive forces between the charges. The problem is most simply analyzed by work-energy conservation, assuming that the work done in expanding the bubble against the external pressure and the surface tension comes from the decrease of potential energy in the electric field outside t bubble. Suppose that the liquid has surface tension coefficient y, that its original radius was a that its final radius is b (after q is addedo and that the atmospheric pressure is p. Show that th charge added to the bubble was q 2Tieeabl(pel3xa ab b5+2Y(a b)] 875 a) Given that CS-EA/d, find the equivalent d/2 capacitance of the combination shown here, in terms of Co, K, Ka, Ks, and K. (Neglect edge effects.) b) Given that Co 8 HP, Ki 3, Km 5, K, 4, Ke 2, find the value of the equivalent capacitance. A/2

Explanation / Answer

(a) In this figure K1 and K2 are in the series and K3 and K4 are in series
Equivalent capacitance of these two will be in parallel.
Let me give them name as C1 , C2 , C3 andC4 respectively having dielectric K1 , K2 , K3 and K4
Now For 1 and 2
1/Cnet 1 = 1/C1 + 1/C2
Cnet1 = C1C2 / C1 + C2
Similarly for 3 and 4
Cnet2 = C3 C4  / C3 + C4
Now this will be in parallel
Cnet = (C1C2 / C1 + C2 ) + (C3 C4  / C3 + C4 )
(b) When Co = 8 uF , K1 = 3
then C1 = KCo = 24 uF
C2 = 40 uF , C3 = 32 uF , C4 = 16 uF
Now putting these value in above equation we get
Cnet = 15 + 10.667 = 25.667 uF

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