Question

An oil drop, carrying six electronic charges and having a mass of $1.6\times {10}^{-12}g$, falls with some terminal velocity in a medium. What magnitude of vertical electric field is required to make the drop move upward with the same speed as it was formerly moving downward with? Ignore buoyancy

• A. ${10}^{5}N{C}^{-1}$
• B. ${10}^{4}N{C}^{-1}$
• C. $3.3\times {10}^{4}N{C}^{-1}$
• D. $3.3\times {10}^{5}N{C}^{-1}$

Answer 1

The correct option is C $3.3\times {10}^{4}N{C}^{-1}$

For the first case, there is only gravitational force to balance drag froce
$F=mg$
For the second case, drop is moving upward so drag force will be in downward direction
$F+mg=6eE$ from the first case we have $F=mg$
$2mg=6eE$
$E=\frac{mg}{3e}=\frac{1.6\times {10}^{-15}\times 10}{3\times 1.6\times {10}^{-19}}=3.3\times {10}^{4}N{C}^{-1}$