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Question:
The electric field required to keep a water drop of mass in just to remain suspended, when charged with one electron:
The electric field required to keep a water drop of mass in just to remain suspended, when charged with one electron:
Updated On: Jul 29, 2022
- $ \text{em/g} $
- $ \text{mg/e} $
- $ \text{emg} $
- $ \text{mg} $
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The Correct Option is B
Solution and Explanation
Here: mass of water drop $ =m, $ so, weight of water drop $ =mg $ so, $ mg= $ electrostatic force applied by the field $ mg=qE=eE\,\,\,\Rightarrow \,\,\,E=\frac{mg}{e} $
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Concepts Used:
Electric Field
Electric Field is the electric force experienced by a unit charge.
The electric force is calculated using the coulomb's law, whose formula is:
\(F=k\dfrac{|q_{1}q_{2}|}{r^{2}}\)
While substituting q2 as 1, electric field becomes:
\(E=k\dfrac{|q_{1}|}{r^{2}}\)
SI unit of Electric Field is V/m (Volt per meter).
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