Eight oil drops of same size are charged to a potential to 50 V each. These oil drops are merged into one single larged drop. What will be the potential of the large drop ?

Options

A. 50 V
B. 100 V
C. 200 V
D. 400 V

Answer

To solve the problem step by step, we will follow the principles of electrostatics and the conservation of charge and volume.

### Step 1: Understand the Given Information
We have 8 oil drops, each charged to a potential of 50 V. We need to find the potential of a single large drop formed by merging these 8 drops.

### Step 2: Conservation of Charge
When the 8 small drops merge into one large drop, the total charge \( Q \) on the large drop is the sum of the charges on the individual drops. If the charge on each small drop is \( q \), then:
\[ Q = 8q \]

### Step 3: Volume Conservation
The volume of the small drops must equal the volume of the large drop. The volume \( V \) of a sphere is given by:
\[ V = \frac{4}{3} \pi r^3 \]
For 8 small drops, the total volume is:
\[ V_{\text{small}} = 8 \times \frac{4}{3} \pi r^3 = \frac{32}{3} \pi r^3 \]
For the large drop with radius \( R \), the volume is:
\[ V_{\text{large}} = \frac{4}{3} \pi R^3 \]
Setting these equal gives:
\[ \frac{32}{3} \pi r^3 = \frac{4}{3} \pi R^3 \]
Cancelling \( \frac{4}{3} \pi \) from both sides, we have:
\[ 8r^3 = R^3 \]
Taking the cube root of both sides:
\[ R = 2r \]

### Step 4: Calculate the Potential of the Large Drop
The potential \( V \) of a charged sphere is given by:
\[ V = \frac{kQ}{R} \]
For the small drop:
\[ V_s = \frac{kq}{r} \]
We know \( V_s = 50 \, \text{V} \), so:
\[ q = V_s \cdot r/k \]
Substituting \( Q = 8q \) into the potential formula for the large drop:
\[ V_b = \frac{k(8q)}{R} = \frac{k(8q)}{2r} \]
Substituting \( q \) from the small drop potential:
\[ V_b = \frac{8}{2} \cdot \frac{kq}{r} = 4 \cdot V_s \]
Thus:
\[ V_b = 4 \cdot 50 \, \text{V} = 200 \, \text{V} \]

### Final Answer
The potential of the large drop is **200 V**.

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