Physics for Scientists and Engineers: A Strategic Approach with Modern Physics (3rd Edition)

The object has a charge of $+1.2\times 10^{-8}~C$
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 $+1.2\times 10^{-8}~C$.