## Physics for Scientists and Engineers: A Strategic Approach with Modern Physics (4th Edition)

The way that we're going to solve this is by looking at the electric field that's coming into the cube, and going out of the cube. We will add up all of the quantities that are going in, and compare that with the sum of the quantities that are exiting the cube. $$E_i=5+20+15=40N/C$$ $$E_o=10+20+10=40N/C$$ Thus, $$E_i=E_o$$ Hence, there is no charge in the box.