The magnitude of the electric field is related to the magnitude of the force experienced by the charge by the relationship $F = |q|E$. To use the equation we have to convert the charge from microcoulombs to coulombs. $1;mumathrm{C} = 1times10^{-6}$ C allows us to do this. Solving for $E$ and substituting the known values, we find

$$
E = frac{F}{|q|} = frac{0.25;mathrm{N}}{2.80times10^{-6};mathrm{C}} = 8.9times10^4;mathrm{N/C}
$$

$$
8.9times10^4;mathrm{N/C}
$$

The weight of the object is given by $W = mtextit{g}$. Hence, the magnitude of the upward electric force is $F = W = mtextit{g}$.

Since the given charge is negative, the electric field and the electric force acting on the charge have opposite directions. As the electric force points upward, the electric field should be directed downward.

The relationship between the electric force and the electric field is given by $F = |q|E$. Expressing $F$ as $mtextit{g}$ gives us $mtextit{g} = |q|E$. Solving the equation for $E$ and substituting the known values (remember $1;mumathrm{C} = 1times10^{-6};mathrm{C}$), we find

$$
E = frac{mtextit{g}}{|q|} = frac{(0.012;mathrm{kg})(9.8;mathrm{m/s}^2)}{3.6times10^{-6};mathrm{C}} = 3.3times10^4;mathrm{N/C}
$$

where we used N = $mathrm{kg}cdotmathrm{m/s}^2$. We see the electric force has a magnitude of $3.3times10^4;mathrm{N/C}$, and it is directed downward.

$3.3times10^4;mathrm{N/C}$         (downward)

Because the electric field and the electric force experienced by the charge are in the same direction, the charge must have a positive sign. The magnitudes of the electric field and the electric force are related by the relationship $F = |q|E$. Solving for $|q|$ and substituting the known values, we find

$$
|q| = frac{F}{E} = frac{0.45;mathrm{N}}{6.1times10^4;mathrm{N/C}} = 7.4;mumathrm{C}
$$

We see the electric charge is $q = +7.4;mu$C.

$+7.4;mu$C