Physics, asked by vedeshsharma221102, 4 months ago

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 yadavjii64
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 gameraryansharma9911
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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