Solveeit Logo

Question

Question: 1000 small water drops each of capacitance C join together to form one large spherical drop. The cap...

1000 small water drops each of capacitance C join together to form one large spherical drop. The capacitance of a bigger sphere is
A. C
B. 10 C
C. 100 C
D. 1000 C

Explanation

Solution

In order to solve the question, we will first find the radius of both the drops after then we will use the electric potential and capacitance relation to find the relation between radius and charge after then we will use the given capacitance of small drop and radius of drops to find the capacitance of big drop.

Formula used:
Electric potential on surface
V=KqrV = \dfrac{{Kq}}{r}
K is Coulomb's Constant
Q is charge
V potential difference
R is radius
Q = CV
C is capacitance
Q is charge
V potential difference

Complete step by step answer:
In question we are given,1000 small water drops to form a large spherical drop
Capacitance of small water drop = C
We have to find capacitance of large spherical drop
Volume of large spherical drop = 1000 ×\timesvolume of small spherical drop ……. (Equation 1)
Let the radius of large spherical drop = R
So, volume of large spherical drop = 43πR3\dfrac{4}{3}\pi {R^3}
Let the radius of large spherical drop = r
So, volume of large spherical drop = 43πr3\dfrac{4}{3}\pi {r^3}
Substituting the values of volume in the formula
43πR3=1000×43πr3\dfrac{4}{3}\pi {R^3} = 1000 \times \dfrac{4}{3}\pi {r^3}
Now removing the same values from both side and we get
R=\root3\of1000 r3R = \root 3 \of {1000{\text{ }}{r^3}}
Solving the cube root, we will get
R = 10 r
We know that electric potential on surface of spherical drop is
V=KqrV = \dfrac{{Kq}}{r}
where k is Coulomb's Constant
Comparing with Q=CV
rearranging the above formula we have
C=qVC = \dfrac{q}{V}
Substituting the value of V
C=qKqrC = \dfrac{q}{{\dfrac{{Kq}}{r}}}
Cutting the q, we get
C=rKC = \dfrac{r}{K}
Hence, we can use this relation to find the capacitance of big drop using k as constant
capacitanceradius = contant\dfrac{{{\text{capacitance}}}}{{{\text{radius}}}}{\text{ = contant}}
Using the above formula for equating both the drops
capacitance of big dropradius of big drop = capacitance of small dropradius of small drop\dfrac{{{\text{capacitance of big drop}}}}{{{\text{radius of big drop}}}}{\text{ = }}\dfrac{{{\text{capacitance of small drop}}}}{{{\text{radius of small drop}}}}
capacitance of big drop10 r = Cr\dfrac{{{\text{capacitance of big drop}}}}{{10{\text{ }}r}}{\text{ = }}\dfrac{C}{{\text{r}}}
capacitance of big drop = 10 C{\text{capacitance of big drop = 10 C}}
Hence, the correct option is B) 10 C.

Note: Many of the people will make the mistake by not assuming radius at first place instead of that using the whole volume but this will make the answer lengthier as there is no relation direct relation between capacitance and volume so instead at in first step, we have to rearrange for radius in end steps .