Question

Eight drops of mercury of equal radii and possessing equal charges combine to form a big drop. Then the capacitance of the bigger drop compared to each individual drop is.....

Answer 1

Capacitance of Smaller drop (say Cs)= 4πεr
where "r" is the radius of the drop and "ε" is the permittivity of space.
__________
FINDING RADIUS OF BIG DROP:
Now 8 drops combine to form a single large drop .
then using volume conservation :
Volume of Big drop = 8 times the volume of smaller drops.
(4/3)πR³ =8× (4/3)πr³
=> R = 2r
here "R" is the radius of Big drop.
__________
=> Capacitance of Big Drop (Say Cb) = 4πεR
= 4πε(2r) = 8πεr .
__________
=> Cb/Cs = 2
=> Capacitance of Big drop is two times than that of smaller drop.
__________
hope it helps !

Answer 2

Answer:

Capacitance of Smaller drop (say Cs)= 4πεr

where "r" is the radius of the drop and "ε" is the permitivity of space.

FINDING RADIUS OF BIG DROP:

Now 8 drops combine to form a single large drop .

then using volume conservation :

Volume of Big drop = 8 times the volume of smaller drops.

(4/3)πR³ =8× (4/3)πr³

=> R = 2r

here "R" is the radius of Big drop.

=> Capacitance of Big Drop (Say Cb) = 4πεR

= 4πε(2r) = 8πεr.

=> Cb/Cs = 2

=> Capacitance of Big drop is two times than that of smaller drop.

hope this helps you....

Mark as brainliest