The potential of a large liquid drop when eight liquid drops are combined is 20 V. Then the potential of each single drop was
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First think of a conducting sphere having a charge q and radius r. The potential at the surface (and internally) is p= kq/r
The volume of the sphere v = (4/3)πr³
By combining 2 such spheres, the new sphere has a charge Q= 2q and volume V = 2v.
It is called doubling.
To find the new radius, R:
V= 2v
4/3πR³ = 2 × (4/3)πr³
R³ = 2r³
R = r ³√2
The potential of the new sphere is :
P = kQ/R = k(2q)/(r ³√2)kq/r = (2/³√2)p
You see that 'doubling' results in the potential increasing by a factor 2/ ³√2. This is always true whatever the value of q.
If rights drops are combined, this is the equivalent to 3 successive 'doubling' as follows:
First combine pairs of charges ( q and p) This produce 4 pairs each with potential (2/³√2)p.
Second Combination pairs of the larger charges (2q and 2q). This will again increase potential by a factor 2/³√2. We now have 2 charges (each 4q) and each with potential.
(2/³√2)(2/³√3)p
Thirdly we combine the two 4q charges. This again increases potential by a factor 2/³√2. We know have one charge (8q) with potential.
(2/³√2)³ × p = (8/2)p = 4p
We are told the final potential is 20V so
4p = 20
p = 5V
Hence, the potential of each single drop was 5V.
The volume of the sphere v = (4/3)πr³
By combining 2 such spheres, the new sphere has a charge Q= 2q and volume V = 2v.
It is called doubling.
To find the new radius, R:
V= 2v
4/3πR³ = 2 × (4/3)πr³
R³ = 2r³
R = r ³√2
The potential of the new sphere is :
P = kQ/R = k(2q)/(r ³√2)kq/r = (2/³√2)p
You see that 'doubling' results in the potential increasing by a factor 2/ ³√2. This is always true whatever the value of q.
If rights drops are combined, this is the equivalent to 3 successive 'doubling' as follows:
First combine pairs of charges ( q and p) This produce 4 pairs each with potential (2/³√2)p.
Second Combination pairs of the larger charges (2q and 2q). This will again increase potential by a factor 2/³√2. We now have 2 charges (each 4q) and each with potential.
(2/³√2)(2/³√3)p
Thirdly we combine the two 4q charges. This again increases potential by a factor 2/³√2. We know have one charge (8q) with potential.
(2/³√2)³ × p = (8/2)p = 4p
We are told the final potential is 20V so
4p = 20
p = 5V
Hence, the potential of each single drop was 5V.
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