In Rutherford\'s famous scattering experiments that led to the planetary model o
ID: 1541923 • Letter: I
Question
In Rutherford's famous scattering experiments that led to the planetary model of the atom, alpha particles (having charges of +2e and masses of 6.64 times 10^-27 kg) were fired toward a gold nucleus with charge + 79e. An alpha particle, initially very far from the gold nucleus, is fired at 2.3 times 10^7 m/s directly toward the nucleus, as in the figure below. How close does the alpha particle get to the gold nucleus before turning around? Assume the gold nucleus remains stationary. (show the steps of your calculation and Indicate units to get credit) Two conductors having net charges of +16.0 mu C and -16.0 mu C have a potential difference of 16.0 V between them. Determine the capacitance of the system. What is the potential difference between the two conductors if the charges on each are increased to +256.0 mu C and -256.0 mu C? (show the steps of your calculation and indicate units to get credit) Given a 2.75 mu F capacitor, a 6.00 mu F capacitor, and a 12.00 V battery, find the charge on each capacitor if you connect them in the following ways. in series across the battery in parallel across the battery (show the steps of your calculation and Indicate units lo get credit) A metal sphere of radius 5.00 cm is initially uncharged. How many electrons would have to be placed on the sphere to produce an electric field of magnitude 1 39 times 10^5 N/C at a point 7.80 cm from the center of the sphere? (show the steps of your calculation and indicate units to get credit) Suppose the conducting spherical shell in the figure below carries a charge of 3.60 nC and that a charge of -1.00 nC is at the center of the sphere. If a = 1.60m and b = 2.40 m, find the electric field (magnitude and direction) at the following. r = 1.50 m r = 2.20 m r = 2.50 m What is the charge distribution on the outer surface of the sphere? What is the charge distribution on the inner surface of the sphere? (show the steps of your calculation and indicate units to get credit)Explanation / Answer
1 )
PE = K ( 2 e X 79 e ) / r
PE = K X 158 X e2 / r
0.5 m V2 = K X 158 X e2 / r
r = 2 K X 158 X e2 / m V2
r = 2 X 9 X 109 X 158 X 1.62 X 10-38 / 6.67 X 10-27 X ( 2.3 X 107 )2
r = 2.06 X 10-14 m
2 )
a )
C = q / V
= 16 X 10-6 / 16
C = 10-6 F
b )
Q = C dV
dV = Q / C
dV = 256 X 10-6 / 10-6
dV = 256 volts
3 )
2.75 uF , 6 uF
a )
1 / Cseries = 1 / 2.75 + 1 / 6
Cseries = 1.88 uF
V = 12 volts
Q = C V
Q = 1.88 X 10-6 X 12
Q = 2.25 X 10-5 C
b )
Cparallel = 2.75 + 6
Cparallel = 8.75 uF
Q = C V
Q = 8.75 X 10-6 X 12
Q = 1.05 X 10-4 C
4 )
E = k Q / r2
Q = E r2 / k
Q = 1.39 X 105 X ( 0.05 )2 / 9 X 109
Q = 3.86 X 10-8 C
Q = n e
n = Q / e
n = 3.86 X 10-8 / 1.6 X 10-19
n = 2.41 X 1011 electrons
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