10. Consider a solid, uniformly charged sphere of radius R = 0.65 m, charge Q1 =
ID: 1766085 • Letter: 1
Question
10. Consider a solid, uniformly charged sphere of radius R = 0.65 m, charge Q1 = 1.00×10-5 C, and mass M = 10.0 kg that is fixed in place. A particle with mass m = 1.00 g and charge Q2 = 1.00×10-7 C is located at a height h = 2.30 m above the surface of the sphere. Calculate the electric potential energy of the system. [ans. 3.05×10-3 J]
11. Calculate the gravitational potential energy of the system. [ans. -2.26×10-13 J]
12. If both masses are in deep space, far from any other gravitational and electric fields, what is the initial acceleration of mass m? Take positive acceleration to be away from the sphere. [ans. 1.03 m/s2]
2 (0,0,0)Explanation / Answer
10.
Electrical Potential Energy, Ve=(Q1Q2) / (4R) .....(1)
Here considering free space = 0
R = radius of sphere + distance = 0.65 + 2.30 = 2.95 m
(Note:Important point here is that while calculating 'R' for a charged sphere in finding Force, Field, Potential etc., the radius of sphere also must be added)
Substituting in (1), Ve = (1×10-5×1×10-7) / (4×8.85×10-12×2.95) = 3.05 × 10-3 J
11.
Gravitational Potential Energy, Vg=(-Gm1m2) / R .....(2)
Here also R = radius of sphere + distance = 0.65 + 2.30 = 2.95 m
Substituting in (2), Vg = (-6.67×10-11×10×1×10-3) / 2.95 = 2.26 × 10-13 J
12.
Electrical force on the particle, Fe= Electrical potential/R
= Ve/R = (3.05 × 10-3) / 2.95 = +1.034× 10-3 N (+sign indicates repulsive force)
Gravitational force on the particle, Fg= Gravitational potential/R
= Vg/R = (-2.26 × 10-13) / 2.95 = -0.766× 10-13 N (-sign indicates attractive force)
Total force on the particle, Ftotal = Fe+Fg = 1.034× 10-3 + (-0.766× 10-13) = 1.034× 10-3 (Repulsive force)
Also we know that Ftotal = mass×acceleration
Therefore, acceleration, a = Ftotal/mass = (1.034× 10-3) / (1×10-3) = 1.034 m/s2
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