A large charged metal sheet is placed in a uniform electric field, perpendicularly to the electric field lines. After placing the sheet into the field, the electric field on the left side of the sheet is E1=5×105 Vm−1 and on the right it is E2=3×105 Vm−1. The sheet experiences a net electric force of 0.08 N. Find the area of one face of the sheet. Assume the external field to remain constant after introducing the large sheet. Use (14πε0)=9×109 Nm2C−2

Let external field be E0 and surface charge density of sheet be σ (including both surfaces). EP=σ2ε0Electric field due to sheetE1=E0−EP.............(i)−E2=E0+EP.............(ii)From Eqs. (i) and (ii), E0=E1−E22=105 Vm−1 and EP=−E1−E22=−4×105 Vm−1 σ2ε0=−4×105 or σ=−8ε0×105Force on sheet:F=qE0 or 0.08=σAE0or 0.08=8ε0×105 A×105or A=10−12ε0=10−12×36×109 π=3.6π×10−2 m2

A large charged metal sheet is placed in a uniform electric field, perpendicularly to the electric field lines. After placing the sheet into the field, the electric field on the left side of the sheet is E1=5×105 Vm−1 and on the right it is E2=3×105 Vm−1. The sheet experiences a net electric force of 0.08 N. Find the area of one face of the sheet. Assume the external field to remain constant after introducing the large sheet. Use (14πε0)=9×109 Nm2C−2

Let external field be E0 and surface charge density of sheet be σ (including both surfaces). EP=σ2ε0Electric field due to sheetE1=E0−EP.............(i)−E2=E0+EP.............(ii)From Eqs. (i) and (ii), E0=E1−E22=105 Vm−1 and EP=−E1−E22=−4×105 Vm−1 σ2ε0=−4×105 or σ=−8ε0×105Force on sheet:F=qE0 or 0.08=σAE0or 0.08=8ε0×105 A×105or A=10−12ε0=10−12×36×109 π=3.6π×10−2 m2