Electricity and magnetism - Electrostatic-11

145

Ranadeep Roy ·2013-06-04 08:49:24

D

229

Dwijaraj Paul Chowdhury ·2013-06-04 09:57:22

Only D

1161

Akash Anand ·2013-06-04 22:30:51

What about option A?? Whether it is correct or wrong? Prove if any :)

145

Ranadeep Roy ·2013-06-04 23:39:34

Initially, Aâ†’q_a,R,V_a=2V
Bâ†’q_b,2R,V_b=3/2V

Therefore V_a(initial)=KR(q_a +q_b2)=2V ...(1)
V_b(initial)=K2R(q_b+q_a)=3V2 ...(2)

Finally, V_b=0
K2R(q_a+q_b')=0
therefore q_b'=-q_a

(1)-(2) :-
Kq_a/2R=V/2

therefore q_a=VR/K

Also Kq_b/2R=V
therefore q_b=2VR/K

so q_aq_b=12

229

Dwijaraj Paul Chowdhury ·2013-06-04 23:46:39

Oh ..yea option a is also correct...
2V=k(q_a+q_{b )}2R-------------(i)

3V/2=k(q_a+q_b)2R---------------(ii)
Dividing (i)/(ii) and then on solving we get....
q_a/q_b=1/2

And q_a^'/q_b^'=-1
Hence option A and D

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Dwijaraj Paul Chowdhury Oops..sorry ..i didn't refresh the page..and didnt notice that it was already answered :P
Upvote·0· Reply ·2013-06-04 23:51:12
Akash Anand Any way ...good work Dwijaraj
Upvote·0· Reply ·2013-06-04 23:55:34
Dwijaraj Paul Chowdhury Thank You sir :)
Upvote·0· Reply ·2013-06-05 04:48:15
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Electrostatic-11