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CGP EDU Academic Team
Published on: September 12, 2026
A charged water drop whose radius is $0.1\,\mu m$ is in equilibrium in an electric field. If charge on it is equal to charge of an electron, then intensity of electric field will be $\left(g = 10\, m s^{-1}\right)$
Text Solution
Verified by ExpertsThe correct answer is:
C
In balance condition $QE = mg = \left(\frac{4}{3} \pi r^{3} \rho \right) g$
⇒ ⇒ $E = \frac{4 \times (3.14)(0.1 \times 10^{-6})^3 \times 10^3 \times 10^3}{3 \times 1.6 \times 10^{-19}}$ $= 262 \ \mathbf{N/C}$
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