Answer
(a) $1.3\times 10^3N/C$
(b) $5.3\times 10^{-2}N$
Work Step by Step
(a) We know that
$\Sigma F_y=QEsin\theta-mg=0$
$\implies E=\frac{mg}{Qsin\theta}$
We plug in the known values to obtain:
$E=\frac{(0.0031)(9.81)}{48\times 10^{-6}sin30.0}$
$E=1.3\times 10^3N/C$
(b) We can find the required tension as follows:
$T=QEcos\theta$
but $E=\frac{mg}{Qsin\theta}$
$\implies T=Qcos\theta(\frac{mg}{Qsin\theta})=\frac{mg}{tan\theta}$
We plug in the known values to obtain:
$T=\frac{(0.0031)(9.81)}{tan30^{\circ}}$
$T=5.3\times 10^{-2}N$
