Question #8e958

Apr 25, 2017

For parallel plates, we know that electric field strength $E$ is:

$E = \frac{\Delta V}{\Delta x} = 98 \cdot {10}^{3} \text{ N/C}$

Balancing forces in the vertical direction:

$E q = m g$

$q = \frac{m g}{E} = \frac{{10}^{- 6} \cdot 9.81}{98 \cdot {10}^{3}} \approx {10}^{- 10} \setminus C$

Apr 25, 2017

${10}^{-} 10 C$

Explanation:

Since the oildrop is at equilibrium, downward weight = upward electric force.

${F}_{G} = {F}_{E}$
$m g = q E$
$m g = q \frac{V}{d}$
$\left(1 \cdot {10}^{- 6}\right) \left(9.81\right) = q \frac{980}{1 \cdot {10}^{- 2}}$
$q = 1.00 \cdot {10}^{- 10} C$