27 identical drops of mercury are charged to the same potential of 10 V.what will be the potential if 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 10 V.what will be the potential if 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 = 10*K(27q)/(27⅓)r.
=(10 * 27)/3 Kq/r
=(10 * 27)/3 V
= 90V
Hope it helps!
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 = 10*K(27q)/(27⅓)r.
=(10 * 27)/3 Kq/r
=(10 * 27)/3 V
= 90V
Hope it helps!
