Answer
(a) $2.1\times 10^4N/C$
(b) $14.4^{\circ}$
Work Step by Step
First, we solve part(b) to find the new angle.
(b) We know that
$T=\frac{mg}{cos\theta}$
$\implies cos\theta=\frac{mg}{T}$
$\implies \theta=cos^{-1}(\frac{mg}{T})$
We plug in the known values to obtain:
$\theta=cos^{-1}(\frac{(0.025Kg)(9.8m/s^2)}{0.253N})$
$\theta=14.4^{\circ}$
(a) We know that
$E=\frac{Tsin\theta}{q}$
We plug in the known values to obtain:
$E=\frac{(0.253N)sin(14.4^{\circ})}{3\times 10^{-6}C}$
$\implies E=2.1\times 10^4N/C$