27 identical drops of mercury are charged to the same potential of 10V . what will be the potential in all the charged drops are made to combine to form one large drop? Assume the drops to be spherical.
27 identical drops of mercury are charged to the same potential of 10V . what will be the potential in all the charged drops are made to combine to form one large drop? Assume the drops to be spherical.
Let potential of each small drop = Kq/r = V (in your case, 10 V).
We know, Volume of n (here 27 small drops) drops = Volume of one big drop.
which implies:
n * 4/3 pi r³ = 4/3 pi R³
where, r is radius of small drop and R is the radius of big drop.
Therefore , R = n⅓ * r and also charge remains conserved.
Therefore, charge on 1 small drop = q and charge on big drop containing n drops = nq .
Therefore,
potential on big drop = K (nq)/ (n⅓)r
= (n⅔) Kq/r
= (n⅔) V
Now substitute the n with the number of drops you wish to combine and do the math.
—
Potential of each small drop = 10V.
No. of drpos : 27.
Potential on big drop = (27)^(2/3)*10
=(27)^(2/3)=(3)^2=9
=(10*9)
= 90
If you are not able to derived it is enough to byheart last equation for your exam point if virw
Let potential of each small drop = Kq/r = V (in your case, 10 V).
We know, Volume of n (here 27 small drops) drops = Volume of one big drop.
which implies:
n * 4/3 pi r³ = 4/3 pi R³
where, r is radius of small drop and R is the radius of big drop.
Therefore , R = n⅓ * r and also charge remains conserved.
Therefore, charge on 1 small drop = q and charge on big drop containing n drops = nq .
Therefore,
potential on big drop = K (nq)/ (n⅓)r
= (n⅔) Kq/r
= (n⅔) V
Now substitute the n with the number of drops you wish to combine and do the math.
—
Potential of each small drop = 10V.
No. of drpos : 27.
Potential on big drop = (27)^(2/3)*10
=(27)^(2/3)=(3)^2=9
=(10*9)
= 90
If you are not able to derived it is enough to byheart last equation for your exam point if virw
