1000 drops of mercury of equal radii and possessing equal charges (each q) combine to form a big drop. What
will be (a) the charge ont he bigger drop and (b) capacitance of bigger frop as compared to capacitance of a
smaller drop?
Answers
Answered by
4
Answer:
Volume of the big drop=Nr(volume of small drop)
If radius of N small drops are 'r',
3
4
(radius of the big drop)×π=N×
3
4
πr
3
Radius of the big drop =N
1/3
r
Capacity (change) of the big spherical drop is,c=4πε
0
R=4πε
0
×N
1/3
×r
Total change of big drop =N
q
(change of each drop)
(capacity)(potential)=change
(4πε
0
N
1/3
r)(v)=N
q
v=
4πε
0
r
N
2/3
q
Answered By
yadav jii64
Answered by
0
Answer:
Volume of the big drop=Nr(volume of small drop)
If radius of N small drops are 'r',
3
4
(radius of the big drop)×π=N×
3
4
πr
3
Radius of the big drop =N
1/3
r
Capacity (change) of the big spherical drop is,c=4πε
0
R=4πε
0
×N
1/3
×r
Total change of big drop =N
q
(change of each drop)
(capacity)(potential)=change
(4πε
0
N
1/3
r)(v)=N
q
v=
4πε
0
r
N
2/3
q
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