Answer
The object has a charge of $+1.2\times 10^{-8}~C$
Work Step by Step
We can find the magnitude of the charge that creates a 270,000 N/C electric field at a point 2.0 cm away:
$E = \frac{k~q}{r^2}$
$q = \frac{E~r^2}{k}$
$q = \frac{(270,000~N/C)(0.020~m)^2}{9.0\times 10^9~N~m^2/C^2}$
$q = 1.2\times 10^{-8}~C$
An electric field points away from a positive charge. Therefore, the object has a charge of $+1.2\times 10^{-8}~C$.
