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https://session.masteringphysics.com/myct itemView?assignmentProblemID-62169021 AP Physics 1 Signed in as Zac Peterson Helo FElectricity Part1 Problem 20.10 previous 1 6 of 8 1next Problem 20.10 A small metal sphere has a mass of 0.16 g and a charge of -26.0 nC It is 10 cn directly above an identical sphere with the same charge. This lower sphere is fixed and cannot move. Part A What is the magnitude of the force between the spheres? Express your answer to two significant figures and include the appropriate units. F Value Units Submit My Answers Give Up Part B If the upper sphere is released, it will begin to fall. What is the magnitude of its initial acceleratian? Express your answer to two significant figures and include the appropriate units. asValue Units Submit My Answers Give Up Continue 901 AMExplanation / Answer
a)
Electric force between 2 charged particles (derived from Coulomb's Law)
F = Q1Q2 / (4r^2)
Q1 = charge on first particle
Q2 = charge on second particle
= permittivity of free space (a constant, actually epsilon nought, can't do subscript here), 8.85 x 10^-12 F/m
r is the distance between the centres of the 2 particles.
So need to convert everything to SI units.
F = (-26n)(-26n) / [(4)(0.1^2)]
= 6.08 x 10^-4 N (3 sig fig)
b) Since charges are negative on both spheres, they experienced a repelling electric force.
Upon release, the sphere is now only under 2 forces, the electric force and its weight. Since it is directly above the lower sphere, the electric force is acting directly upwards, hence negating some of the weight.
Net force on sphere = Weight - Fe (taking downward to be positive)
Weight = mg
= 0.16 x 10^-3 x 9.81
= 1.56 x 10^-3N
Net force on sphere = (1.56 x 10^-3) - (6.08 x 10^-4)
= 9.52 x 10^-4 (Hence net downward force, therefore begins to fall)
Initial acceleration under this force
Fnet = ma
a =9.52 x 10^-4 / (0.16 x 10^-3)
= 5.95m/s^2
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