Question

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

• A. $9.8\times {10}^{-16}$
• B. $9.8\times {10}^{16}$
• C. $9.8\times {10}^{-13}$
• D. $9.8\times {10}^{13}$
• E. $9.8\times {10}^{10}$

Answer 1

Answer: (D)

$9.8\times {10}^{13}$

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