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 6.0R1 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

a)The potential will be equal

V1 = V2

b)The charge will continue to flow until both the spheres are at the same potential.

Q1/R1 = Q2/6R1

Q2 = 6Q1

Q = Q1 + Q2 = Q1 + 6Q1 = 7Q1 =>Q1 = Q/7

Hence, Q1 = Q/7

c)Q2 = 6Q1 = 6 x Q/6

Hence, Q2 = 6/7 Q

d)we know, sigma = Q/A

sigma1 = Q1/A1 = Q/7 x (4 pi R1^2)

sigma2 = Q2/A2 = 6Q/7 x 4 pi (6R2)^2 = Q/42 (4 pi R1^2)

sigma1/sigma2 = Q/7 (4 pi R1^2) x 42 (4 pi R1^2)/Q = 6.00

Hence, sigma1/sigma2 = 6.00

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