Question

A point change is placed at the center of a hollow conducting sphere of internal radius r and outer radius 2r. the ratio of the surface charge density of the inner surface to that of the outer surface will be:
1 : 2
2 : 1
4 : 1
1 : 4

Answer 1

Answer:

its answer is 4:1

Explanation:

surface charge density of inner surface = q/4pie×r^2

and surface charge density of outer surface will be= q/4pie×(2r)^2

their ratio = (q/4pie×r^2) / (q/4pie×4r^2)= 4:1

Answer 2

Answer

here is ur ans

Step by step explanation

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