Question:
The electric field strength in \( \text{N C}^{-1} \) that is required to just prevent a water drop carrying a charge \( 1.6 \times 10^{-19} \) C from falling under gravity is (\( g = 9.8 \, \text{m s}^{-2} \), mass of water drop = \( 0.0016 \) g)
The electric field strength in \( \text{N C}^{-1} \) that is required to just prevent a water drop carrying a charge \( 1.6 \times 10^{-19} \) C from falling under gravity is (\( g = 9.8 \, \text{m s}^{-2} \), mass of water drop = \( 0.0016 \) g)
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Equilibrium → set electric force equal to weight.
Updated On: May 2, 2026
- $9.8 \times 10^{-16}$
- $9.8 \times 10^{16}$
- $9.8 \times 10^{-13}$
- $9.8 \times 10^{13}$
- $9.8 \times 10^{10}$
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The Correct Option is
Solution and Explanation
Concept:
To just prevent falling:
\[
\text{Electric force} = \text{Gravitational force}
\]
\[
qE = mg
\]
Step 1: Convert mass: \[ 0.0016 \text{ g} = 1.6 \times 10^{-6} \text{ kg} \]
Step 2: Substitute: \[ E = \frac{mg}{q} \] \[ E = \frac{1.6 \times 10^{-6} \times 9.8}{1.6 \times 10^{-19}} \]
Step 3: Simplify: \[ E = 9.8 \times 10^{13} \] Final Answer: \[ 9.8 \times 10^{13} \, \text{N/C} \]
Step 1: Convert mass: \[ 0.0016 \text{ g} = 1.6 \times 10^{-6} \text{ kg} \]
Step 2: Substitute: \[ E = \frac{mg}{q} \] \[ E = \frac{1.6 \times 10^{-6} \times 9.8}{1.6 \times 10^{-19}} \]
Step 3: Simplify: \[ E = 9.8 \times 10^{13} \] Final Answer: \[ 9.8 \times 10^{13} \, \text{N/C} \]
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