Sphere 1 with radius R 1 has positive charge q . Sphere 2 with radius 4.0 R 1 is

Question

Sphere 1 with radius R1 has positive charge q. Sphere 2 with radius 4.0R1 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 ofq 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) after connection, flow of charges stops when potential across wire becomes
equal.

V1 = V2 ( equal to )

b) so kq1 / r1 = kq2 / r2

q1 / R1 = q2 / 4R1

q1 / q2 = 1/4

b) q1 / (q1 + q2) = (q1 / q2) / ( q1/q2 + 1 )

= 0.25 / 1.25 = 0.2

c) q2 / (q1 + q2) = 1 / (q1/q2 + 1) = 1 / 1.25 = 0.8

d) ratio of surface charges densities :

sigma1 / sigma2 = [ 0.2q / 4pi R1^2 ] / [ 0.8q / 4 pi (4R1)^2 ]

= 0.2 / ( 0.8 / 16) = 4

Related Questions

Navigate

• Browse (All)
• Browse S
• Subjects

Integrity-first tutoring: explanations and feedback only — we do not complete graded work. Learn more.