Sphere 1 with radius R_1 has positive charge q. Sphere 2 with radius 5.0R_1 is f

Question

Sphere 1 with radius R_1 has positive charge q. Sphere 2 with radius 5.0R_1 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.) ________ Number ________ Units ______ Number ________ Units ______ Number ________ Units ______

Explanation / Answer

find the ratio of the charge, think in this way. A bigger sphere has a bigger surface area, therefore, it is able to hold more charge as compare to a smaller sphere.

to find charge of sphere 1 = q* r1/(r1+r2)

to find charge of sphere 2 = q* r2/(r1+r2)

R/(1+5) = R/6

(a) Potential v1 of sphere 1 = Potential v2 of sphere 2.

(b) sphere 1
  
5/6 C/m

(c) sphere 2

1/6 C/m

(d) charge density = q / r^2

sphere 1 : sphere 2

(1/6)/r^2 : (5/6)/(5r)^2

(1/6)/r^2 : (5/6)/25r^2

1/6*r^2 : 1/30*r^2

mutiple by 6*r^2

1 : 1/5

1 C/m : 0.2 C/m
  

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