Question:
A charge of \( 10^{-9} \, \text{C} \) is placed on each of the 64 identical drops of radius 2 cm. They are then combined to form a bigger drop. Its potential will be:
A charge of \( 10^{-9} \, \text{C} \) is placed on each of the 64 identical drops of radius 2 cm. They are then combined to form a bigger drop. Its potential will be:
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The potential of a drop is directly proportional to the charge and inversely proportional to the radius. When drops combine, their total volume (and hence their radius) increases, but the total charge remains constant.
Updated On: Apr 22, 2026
- \( 7.2 \times 10^{3} \, \text{V} \)
- \( 7.2 \times 10^{2} \, \text{V} \)
- \( 1.44 \times 10^{2} \, \text{V} \)
- \( 1.44 \times 10^{3} \, \text{V} \)
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The Correct Option is A
Solution and Explanation
Step 1: Understand the formula for the potential of a drop.
The potential \( V \) of a charged drop is given by the formula: \[ V = \frac{kQ}{r} \] where \( k \) is the Coulomb constant (\( k = 9 \times 10^9 \, \text{N m}^2/\text{C}^2 \)), \( Q \) is the charge, and \( r \) is the radius of the drop.
Step 2: Combine the charges.
When \( 64 \) drops of radius \( r \) are combined, the total charge \( Q_{\text{total}} \) remains the same, but the radius \( r_{\text{new}} \) of the new drop increases. The new radius can be calculated using the formula for volume conservation: \[ V_{\text{total}} = 64V_{\text{single}} \] which leads to: \[ r_{\text{new}} = r \times \sqrt[3]{64} = 2 \times 4 = 8 \, \text{cm} \] The new radius is \( 8 \, \text{cm} \).
Step 3: Calculate the potential of the new drop.
Now, substitute the new radius and total charge into the potential formula: \[ V_{\text{new}} = \frac{9 \times 10^9 \times 10^{-9}}{0.08} = 7.2 \times 10^{3} \, \text{V} \]
The potential \( V \) of a charged drop is given by the formula: \[ V = \frac{kQ}{r} \] where \( k \) is the Coulomb constant (\( k = 9 \times 10^9 \, \text{N m}^2/\text{C}^2 \)), \( Q \) is the charge, and \( r \) is the radius of the drop.
Step 2: Combine the charges.
When \( 64 \) drops of radius \( r \) are combined, the total charge \( Q_{\text{total}} \) remains the same, but the radius \( r_{\text{new}} \) of the new drop increases. The new radius can be calculated using the formula for volume conservation: \[ V_{\text{total}} = 64V_{\text{single}} \] which leads to: \[ r_{\text{new}} = r \times \sqrt[3]{64} = 2 \times 4 = 8 \, \text{cm} \] The new radius is \( 8 \, \text{cm} \).
Step 3: Calculate the potential of the new drop.
Now, substitute the new radius and total charge into the potential formula: \[ V_{\text{new}} = \frac{9 \times 10^9 \times 10^{-9}}{0.08} = 7.2 \times 10^{3} \, \text{V} \]
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