Sphere 1 with radius R 1 has positive charge q . Sphere 2 with radius 6.0 R 1 is

Question

Sphere 1 with radius R1 has positive charge q. Sphere 2 with radius 6.0 R1 is far from sphere 1 and initially uncharged. After the separated spheres are connected with a wire thin enough to retain only negligible charge, (a) is potential V1 of sphere 1 greater than, less than, or equal to potential V2 of sphere 2? What fraction of q ends up on(b) sphere 1 and (c) sphere 2? (d) What is the ratio of the final surface charge density of sphere 1 to that of sphere 2? (Answer with 2 significant figures.)

Explanation / Answer

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Sphere 1 with radius R1 has positive charge q. Sphere 2 with radius 7.00R1 is far from sphere 1 and initially uncharged. The separated spheres are then connected with a wire thin enough to retain only negligible charge.
a.) What is the ratio V1/V2 of the final potentials of the spheres?
b.) What fraction of q ends up on sphere 1?
c.) What fraction of q ends up on sphere 2?
d.) What is the ratio s1/s2 of the surface charge densities of the spheres?

solution:

since both are connected by same wire both should be at same potential
V1=V2
V1/V2 =1

b) V1= V2
q1/c1 =q2/c2
q1/4??R =q2/4??(7R)

7 q1=q2

q1+q2 =q

8q1 =q

q1=q/8 fraction =1/8

c) q2 =7q/8 fraction =7/8

d)?1/?2 = q1*(7R)2/q2R2 = 49/7 = 7

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