Answer
(a) $2.6\times 10^4N/C$
(b) $0.066N$
Work Step by Step
(a) We can find the required electric field as
$E=\frac{Kq}{r^2}$
We plug in the known values to obtain:
$E=\frac{(8.99\times 10^9)(4.2\times 10^{-6})}{(1.2)^2}$
$E=2.6\times 10^4N/C$
(b) We can find the tension in the string as follows:
$T=mg-F=mg-qE$
We plug in the known values to obtain:
$T=(0.0150)(9.81)-(3.1\times 10^{-6})(2.6\times 10^4)$
$T=0.147-0.081=0.066N$
