Sphere 1 with radius R1 has positive charge q. Sphere 2 with radius 8.0R1 is far

Question

Sphere 1 with radius R1 has positive charge q. Sphere 2 with radius 8.0R1 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

Just prior to connecting the spheres, Sphere 1 is at a higher potential than Sphere 2. Connecting them brings them to the same potential - if they weren't at the same potential, charge would flow between them until they reached the same potential.

A. V1 = V2

The potentials are given by:

V1 = kq1/R1 V2 = V1 = kq2/R2 = kq2/(8R1) and conservation of charge requires Q = q1 + q2

B. q1/R1 = q2/(8R1) --> q1 = q2/8 --> Q = q1 + q2 = q2/8 + q2 = (9/8) q2 --> q2 = (8/9)Q and q1 = Q/9

C. Surface charge density s = q/A and for a sphere A = 4*pi*r^2 so

s1 = Q*4*pi*R1^2/9 s2 = Q*(8/9)*4*pi*(8R1)^2 --> s1/s2 = 1/(8*64) = 0.00195

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