Answer
(a) $2.35\times 10^{-6}$
(b) $E = 4.26\times 10^7~N/C$
The electric field must be directed upward.
Work Step by Step
(a) We can find the ratio of the electric force to the bee's weight:
$\frac{q~E}{mg} = \frac{(23\times 10^{-12}~C)(100~N/C)}{(0.10\times 10^{-3}~kg)(9.8~m/s^2)} = 2.35\times 10^{-6}$
(b) We can find the magnitude of the required electric field:
$q~E = mg$
$E = \frac{mg}{q}$
$E = \frac{(0.10\times 10^{-3}~kg)(9.8~m/s^2)}{23\times 10^{-12}~C}$
$E = 4.26\times 10^7~N/C$
Since the charge on the bee is positive, and the direction of the force due to the electric field must be upward, the electric field must be directed upward.
