1. . -13 points My Notes Two conducting spheres with diameters of 0.200 m and 1.
ID: 1874447 • Letter: 1
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1. . -13 points My Notes Two conducting spheres with diameters of 0.200 m and 1.00 m are separated by a distance that is large compared with the diameters. The spheres are connected by a thin wire and are charged to 5.00 Auc. (a) How is this total charge shared between the spheres? (Ignore any charge on the wire.) charge on smaller sphere Auc charge on larger sphere Auc (b) What is the potential of the system of spheres when the reference potential is taken to be V = 0 at r = ? KV 2. - 11 points My Notes A 1 megabit computer memory chip contains many 55.0 FF capacitors. Each capacitor has a plate area of 20.0x 10-12 m2. Determine the plate separation of such a capacitor (assume a parallel-plate configuration). The order of magnitude of the diameter of an atom is 10-10 m = 0.100 nm. Express the plate separation in nanometers. nm 3. O -13 points IMy Notes Consider the circuit shown in Figure P26.23, where C1 = 2.00 AUF, C2 = 1.00 AUF, and AV = 19.0 V. Capacitor C, is first charged by the closing of switch S1. Switch S1 is then opened, and the charged capacitor is connected to the uncharged capacitor by the closing of Sz. Calculate the initial charge acquired by C. Âu. Calculate the final charge on each capacitor. C1 = C2 = ÅuC Figure P26.23Explanation / Answer
1. (A) as they are connected by wire so they are at same potential .
V = k q / r = k Q / R
q / (0.2/2) = Q / (1/2)
Q = 5q
and Q + q = 5 uC
5 q + q = 5
q = 0.833 uC ......smaller sphere
and Q = 4.167 uC .....Larger sphere
(B) V = (9 x 10^9)(0.833 x 10^-6)/(0.2/2)
V = 75000 Volt or 75 kV
3. initial charge, Qi = C1 V = 2uF x 19
= 38 uC ...Ans
after that they are connected in parallel so they wil have same voltage across them.
Applying charge conservation,
C1 V + C2 V = 38
V = 38 uC / (2 + 1)uF = 12.7 Volt
C1 charge = C1 V = 25.3 uC ....Ans
C2 charge = 12.7 uC ....Ans
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