The electric field in a region is given by E→=E0xli→ Find the charge contained inside a cubical volume bounded by the surfaces x=0,  x=a,  y=0,  y=a,  z=0  and  z=a. Take E0=5×103 NC−1,  l=0.02  m,  and  a=0.01  m.

GivenE→=E0li^,  l=2  cm,  a=1  cm,  E0=5×103  NC−1We see that flux passes mainly through surface area ABDC and EFGH. As the AEFB and CHGD are parallel to the flux again in ABDC=0, the flux only passes through the surface area EFGHE = 0.Flux =E0al×area=5×103×al×a2=5×103×a3l=5×103×(0.01)32×10−2=2.5×10−1So, q=ε0×flux=8.85×10−12×2.5×10−1=22.125×10−13=2.2125×10−12  C