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Question 1: You measured the force between two charged pith balls where the pith balls were of the same size and mass. Your assumption was that the charge on the balls was the same.

Consider this change to that situation. Instead of identical pith balls, one has twice the charge of the other. Both pith balls hang from strings of length L. Assume the masses are the same and both pith balls hang at the same angle. Derive an expression for the distance between the pith balls.

2)

0.345 m and uniform charge density 151 nC/m3 lies at the center of a spherical, conducting shell of inner and outer radii 3.50R and 4.00R, respectively. If the conducting shell carries a total charge of Q - 40.5 nC, find the magnitude of the electric field at the given radial distances from the center of the charge distribution. 0.560R: N/C 3.55R: N/C 1.10R: N/C 5.70R: N/C 3.50R 4.00R

Answer 1

Question 1 The free body diagram is as follows. L. T cose r sin e 2q mg Let r be the distance between the two balls From the above diagram, the forces are, T sin 6 F T cos-mg Divide the above two equations,

Tsin θ F, F. tan mg mg 2kq mgr Rearrange the above equation for the dista 2kg mgr 2kg mg tan θ 2kq Ving tan θ 2) At 0.560R: The charge density of the sphere is,

9 σ= 4 3 π(0.560R) 01o) ))3) =(-151x 10-9 C/m3): π(0.560 (0.345 m 4.56x10-9 C The electric field at 0.560R is,

9×109)(-4.56×10-9 C (0.560 (0.345 m)) --1099.49 N/C At 3.505R: The radius is in between the given radius. So, the electric field is zero inside the conducting shell. So, the electric field at 3.505R is zero At 1.10R: The charge density of the sphere is, (-151x10-cin')( π(1.10(0.345 m))3 =-3.46x10-8 C

The electric field at 1.10R is, 9x10" )(-3.46x10- C (1.10(0.345m) =-2162.19 NC At 5.70R The charge density of the sphere is, (-151×10-9 C/㎡)| --πί5·70(0.345 m)) -4.81x10-5 C The electric field at 5.70R is,

9%109)(-4.81x10-6 C+40.5x10° C (5.70 (0.345 m) -11.1x103 NIC