Electrostatics 1 Question 13

13. For a=0, the value of d (maximum value of pas shown in the figure) is

(a) 3Ze4πR3

(b) 3ZeπR3

(c) 4Ze3πR3

(d) Ze3πR3

(2008,4 M)

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Answer:

Correct Answer: 13. (b)

Solution:

  1. For a=0ρ(r)=dRr+d

Now

0R(4πr2)ddRrdr= net charge = Ze. 

Solving this equation, we get d=3ZeπR3



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