Correct Answer - Option 2 : 5 V
The correct answer is option 2) i.e. 5 V.
CONCEPT:
-
Electric potential: The electric potential is also known as voltage (V), is the difference in potential energy per unit charge between two points in an electric field.
The electric potential at any point around a point charge is given as:
\(V = \frac{kQ}{r}\)
Where Q is the point charge, r is the distance from the point charge and k is the constant with a value equal to 9\(\times\)109 Nm2/C2.
-
Principle of conservation of charge: Conservation of charge is based on the principle that the total electric charge in an isolated system remains constant.
CALCULATION:
Given that:
The potential of the large liquid drop (VL) = 20 V
| |
Small Drop |
Large Drop |
| Potential |
VS
|
VL
|
| Charge |
q |
Q |
| Radius |
r |
R |
From the principle of conservation of charge,
Potential due to 8 small drops = Potential due to large drop
8\(\times\)VS = VL
8\(\times\)q = Q ----(1)
Since the large drop is combined from 8 drops,
Volumes of 8 drops = volume of a large drop
\(8\times\frac{4}{3}\:\pi r^3 = \frac{4}{3}\:\pi R^3 \)
\(\Rightarrow 2r =R\) ----(2)
From (1),
\(V_S = \frac{kq}{r}\) and \(V_L = \frac{kQ}{R}\)
\(\frac {V_S}{V_L} = \frac{\frac{8kq}{r}}{\frac{kQ}{R}} = \frac {q}{Q} \times \frac{R}{r}\)
\(\frac {V_S}{20} = \frac {q}{8q} \times \frac{2r}{r} = \frac{1}{4}\) (Substituting 2r = R)
The potential of a single drop, VS = 5 V
