Consider two finite thin insulating square sheets, each having a positive surface charge density σ . Their common edge is along x-axis and the sheets are perpendicular to each other.The magnitude of electric field at the location (a2,a2,a2) is √2nσε0 . Find n.
Question
Consider two finite thin insulating square sheets, each having a positive surface charge density σ . Their common edge is along x-axis and the sheets are perpendicular to each other.The magnitude of electric field at the location (a2,a2,a2) is √2nσε0 . Find n.
Solution
The electric field due to a uniformly charged infinite plane sheet is given by E = σ/2ε0, where σ is the surface charge density and ε0 is the permittivity of free space.
Since the sheets are perpendicular to each other, the electric fields due to each sheet will be at right angles to each other.
Let's consider the electric field due to one sheet. At the location (a2,a2,a2), the electric field due to this sheet will be E1 = σ/2ε0.
Similarly, the electric field due to the other sheet will also be E2 = σ/2ε0.
Since the electric fields are at right angles to each other, the resultant electric field E will be given by the vector sum of E1 and E2.
Using the Pythagorean theorem, we get E = √(E1^2 + E2^2) = √[(σ/2ε0)^2 + (σ/2ε0)^2] = σ/2ε0 * √2.
Given that E = √2nσε0, we can equate the two expressions for E and solve for n:
σ/2ε0 * √2 = √2nσε0 => n = 1/4.
So, n = 1/4.
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