All Solutions
Page 708: Practice Problems
$$
E = frac{F}{|q|} = frac{0.25;mathrm{N}}{2.80times10^{-6};mathrm{C}} = 8.9times10^4;mathrm{N/C}
$$
8.9times10^4;mathrm{N/C}
$$
Since the given charge is negative, the electric field and the electric force acting on the charge have opposite directions. As the electric force points upward, the electric field should be directed downward.
The relationship between the electric force and the electric field is given by $F = |q|E$. Expressing $F$ as $mtextit{g}$ gives us $mtextit{g} = |q|E$. Solving the equation for $E$ and substituting the known values (remember $1;mumathrm{C} = 1times10^{-6};mathrm{C}$), we find
$$
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}
$$
where we used N = $mathrm{kg}cdotmathrm{m/s}^2$. We see the electric force has a magnitude of $3.3times10^4;mathrm{N/C}$, and it is directed downward.
$$
|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.