Question

27 spherical drops of same capacitance c are merged to form a big spherical drop. Capacitance of bigger drop is

Answer 1

Let me derive the general expressions for this condition.
Let there be
n
small drops each having a charge
q
on it and the radius
r
,
V
be its potential and let the volume of each be denoted by
B
.
When these
n
small drops are coalesced there is a new bigger drop formed.
Let the radius of the bigger drop be
R
,
Q
be charge on it,
V
'
be its potential and its volume be
B
'
The volume of the bigger drop must be equal to the sum of volumes of
n
individual drops.
⇒
B
'
=
B
+
B
+
B
+
...
...
+
B
There are total
n
small drops therefore the sum of volumes of all the individual drops must be
n
B
.
⇒
B
'
=
n
B
A drop is spherical in shape. Volume of a sphere is given by
4
3
π
r
3
where
r
is its radius.
⇒
4
3
π
R
3
=
n
4
3
π
r
3
⇒
R
3
=
n
r
3
Taking third root on both sides.
⇒
R
=
n
1
3
r
Also the charge of the bigger drop must be equal to the sum of charges on the individual drops.
⇒
Q
=
n
q
The potential of the bigger drop can be given by
V
'
=
k
Q
R
⇒
V
'
=
k
n
q
n
1
3
r
⇒
V
'
=
n
1
−
1
3
k
q
r
⇒
V
'
=
n
2
3
k
q
r
Since,
k
q
r
represents the potential of small drop which we have symbolized by
V
.
Therefore,
V
'
=
n
2
3
V
Now we have found a general equation for this case.
In this case there are
27
identical drops.
⇒
V
'
=
27
2
3
V
⇒
V
'
=
9
V
This shows that in your case the potential of the bigger drop is
9
times the potential of the smaller drop.

Answer 2

Answer:

capacitance of bigger drop is 3 times the smaller drop.

Step-by-step explanation:

Let r be the radius of each small drop and R be the radius of bigger drop.

Now, since 27 spherical drops of same capacitance c are merged to form a big spherical drop.

Thus, we have

Volume of bigger drop = 27× volume of smaller drop

$\frac{4}{3}\pi R^3=27\times\frac{4}{3}\pi r^3\\\\R^3=27\times r^3\\\\R=3r...(i)$

Now, we know the relation between charge, electric potential and capacitance which is given by

q = cv

Also, the potential is given by (small drop)

$v=\frac{kq}{r}$

Substituting the value of q

$v=\frac{kcv}{r}\\\\c=\frac{r}{k}$

Let c' be the capacitance of the bigger drop.  Hence, for bigger drop

$c'=\frac{R}{k}$

From equation (i), substituting the value of R

$c'=\frac{3r}{k}$

Now, r/k = c . Thus, we have

$c'=3c$

Therefore, capacitance of bigger drop is 3 times the smaller drop.