Answer in units of N the distance Op 2a 014 (part 1 of 2) 10.0 points Two conduc
ID: 1868891 • Letter: A
Question
Answer in units of N the distance Op 2a 014 (part 1 of 2) 10.0 points Two conducting spheres have identical radii Initially they have charg's of opposite sign and unequal magnitudes with the magnitude of the positive charge larger than the mag- nitude of the negative charge. They attract each other with a force of 0.155 N when sepa- rated by 0.3 m 2.E= 3.E; 4. E. Eg Initial Thespletes are sudenly connected bya thin conducting wire, which is then removed 7.E. Connected 017 (part 1 of 2) 10.0 points Now the spheres repel each other with a force f 0.033 N An electron and a proton are each placed at rest in an electric feld of 6o1 N/C What is the velocity of the electron 32 2 ns after being released? Consider the direction parallel to the field to be positive. The fun- damental charge is 1.602 x 10-19 C and the Final What is the magnitude of the positive mass of an electron is 9.109 x 10 Answer in units of C. kg charge? Answer in units of m/s 018 (part 2 of 2) 10.0 points What is the velocity of the proton 32.2 after being released? The mass of the proton is 1.6726 × 10-27 kg. 015 (part 2 of 2) 10.0 points What is the negative charge? Answer in units of C Answer in units of m/s 16 10.0 points Consider two concentric spherical conducting 019 10.0 points shells. The inner shell has radius a and charge A4piece of Styrofoam carries a net charge 91 on it, while the outer shell has radius 3a of-0.5 C and floats above the center of a very large horizontal sheet of plastic that has and charge g on a uniform charge density on its surface. What is the charge per unit area on the plastic sheet? The acceleration due to gravity is 9.8 m/s and the permittivity of free space is 8.85419 x 10-12 C2/N/m Answer in units of m:/ 020 10.0 points Two electrons in an atom are separated by 5 x 10-10m, the typical size of an atom Determine the electric field E at p, where What is the force between them? TheExplanation / Answer
Initial
charge on left sphere = +q1
charge on right sphere = -q2
total charge = q1 - q2
force F1 = k*q1*q2/r^2 = 9*10^9*q1*q2/0.3^2 = 0.155
q1*q2 = 1.55*10^-12
final
charge on left sphere q1' = +q
charge on right sphere q2' = +q
total charge = q1' + q2'
force F1 = k*q^2/r^2 = 9*10^9*q^2/0.3^2 = 0.033
q = 5.74*10^-7 C
total charge q + q = 2q = q1 - q2
q1 - q2 = 11.48*10^-7
q2 = q1 - 11.48*10^-7
q1*q2 = 1.55*10^-12
q1*(q1 -11.48*10^-7) = 1.55*10^-12
q1 = 1.94*10^-6 C
q2 = q1 - 11.48*10^-7 = 0.792*10^-6 C
part(a)
magnitude of positive charge = 1.94*10^-6 C
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part(b)
negative charge = -0.792*10^-6 C
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016)
E = k*q1/r1^2
r1 = op = 2a
E = q1/(4*pi*e0*(2a)^2)
E = q1/(16*pi*e0*a^2)
OPTION (3)
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(017)
electric force F = -E*q
acceleration a = F/m = -Eq/m
v = vo + a*t
v = 0 - E*q*t/m
v = -601*1.602*10^-19*32.2*10^-9/(9.109*10^-31)
v = -3.4*10^7 m/s
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018
electric force F = E*q
acceleration a = F/m = Eq/m
v = vo + a*t
v = 0 + E*q*t/m
v = 601*1.602*10^-19*32.2*10^-9/(1.6726*10^-27)
v = 1853 m/s
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