Answer
(a) $0.044Kg $
(b) $6.058\times 10^4N/C $
Work Step by Step
(a) The mass of the object can be determined as follows:
$ m=\frac{Tcos\theta}{g}$
We plug in the known values to obtain:
$ m=\frac{(0.450N)(0.96)}{9.8m/s^2}$
$ m=0.044Kg $
(b) The required electric field can be determined as
$ E=\frac{Tsin\theta}{q}$
We plug in the known values to obtain:
$ E=\frac{(0.450N)(sin16^{\circ})}{2.05\times 10^{-6}C}$
$ E=6.058\times 10^4N/C $
