- Home
- Standard 12
- Physics
1. Electric Charges and Fields
medium
An oil drop carries six electronic charges, has a mass of $1.6 \times 10^{-12} g$ and falls with a terminal velocity in air. The magnitude of vertical electrical electric field required to make the drop move upward with the same speed as was formely moving is ........$kN/C$
A
$16.3$
B
$32.7$
C
$98$
D
None of the above
Solution
For the first case, there is only gravitational force to balance drag froce $F=m g$
For the second case, drop is moving upward so drag force will be in downward direction $F+m g=6 e E$ from the first case we have $F=m g$
$2 m g=6 e E$
$E=\frac{m g}{3 e}=\frac{1.6 \times 10^{-15} \times 10}{3 \times 1.6 \times 10^{-19}}=3.27 \times 10^{4} \mathrm{NC}^{-1}$
Standard 12
Physics