Question:

The electric field strength in $N \,C\,^{-1}$ that is required to just prevent a water drop carrying a charge $1.6 \times 10^{-19} \, C$ from falling under gravity is $(g\, = \,9.8\, ms^{-2}$, mass of water drop = $0.0016\,g$)

Updated On: Jul 29, 2022
  • $9.8 \times 10^{16}$
  • $9.8 \times 10^{-16}$
  • $9.8\times 10^{-13}$
  • $9.8\times 10^{13}$
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The Correct Option is D

Solution and Explanation

Charge on drop $=1.6 \times 10^{-19} C$ Mass of the drop $=0.0016\, g$ $=16 \times 10^{-4}\, g $ $=16 \times 10^{-7}\, kg $ $g=9.8\, m / s ^{-2}$ In balance position, $m g=q E$ where, $E$ is electric field strength. $E =\frac{m g}{q}=\frac{16 \times 10^{-7} \times 9.8}{1.6 \times 10^{-19}}$ $=9.8 \times 10^{13} N / C$
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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).