Two balls with equal charge are in a vessel with ice at - `10 ^@C` at a distance of 25 cm from each other. On forming water at `0^@C` , the balls are brought nearer to 5 cm for the interaction between them to be same. If the dielectric constant of water at `0^@C ` is 80 , the dielectric constant of ice at `-10^@C` is

Options

A. 40
B. `3.2`
C. 20
D. `6.4`

Answer

To solve the problem step by step, we need to analyze the forces acting on the two charged balls in two different scenarios: when they are in ice at -10°C and when they are in water at 0°C.

### Step 1: Understand the Forces in Both Scenarios
The force between two charges \( Q \) and \( Q \) separated by a distance \( r \) in a medium with dielectric constant \( k \) is given by:
\[ F = \frac{1}{4 \pi \epsilon_0} \frac{Q^2}{k r^2} \]
where \( \epsilon_0 \) is the permittivity of free space.

### Step 2: Set Up the Equation for the First Scenario (Ice at -10°C)
In the first scenario, the distance between the charges is 25 cm (0.25 m) and the dielectric constant is \( k \):
\[ F_1 = \frac{1}{4 \pi \epsilon_0} \frac{Q^2}{k (0.25)^2} \]

### Step 3: Set Up the Equation for the Second Scenario (Water at 0°C)
In the second scenario, the distance between the charges is 5 cm (0.05 m) and the dielectric constant is \( k' = 80 \):
\[ F_2 = \frac{1}{4 \pi \epsilon_0} \frac{Q^2}{80 (0.05)^2} \]

### Step 4: Set the Forces Equal
Since the problem states that the interaction between the two balls is the same in both scenarios, we can set \( F_1 = F_2 \):
\[ \frac{1}{4 \pi \epsilon_0} \frac{Q^2}{k (0.25)^2} = \frac{1}{4 \pi \epsilon_0} \frac{Q^2}{80 (0.05)^2} \]

### Step 5: Cancel Common Terms
We can cancel out \( \frac{1}{4 \pi \epsilon_0} \) and \( Q^2 \) from both sides:
\[ \frac{1}{k (0.25)^2} = \frac{1}{80 (0.05)^2} \]

### Step 6: Simplify the Equation
This simplifies to:
\[ \frac{1}{k \cdot 0.0625} = \frac{1}{80 \cdot 0.0025} \]
Calculating \( 80 \cdot 0.0025 \):
\[ 80 \cdot 0.0025 = 0.2 \]
Thus, we have:
\[ \frac{1}{k \cdot 0.0625} = \frac{1}{0.2} \]

### Step 7: Cross Multiply
Cross multiplying gives:
\[ 0.2 = k \cdot 0.0625 \]

### Step 8: Solve for \( k \)
Now, solving for \( k \):
\[ k = \frac{0.2}{0.0625} = \frac{20}{6.25} = 3.2 \]

### Conclusion
The dielectric constant of ice at -10°C is \( k = 3.2 \).

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