Acharge q is uniformly spread over one surface of a very large nonconducting square elastic sheet having sides of length d. at a point p that is 1.25 cm outside the sheet, the magnitude of the electric field due to the sheet is e. if the sheet is now stretched so that its sides have length 2 d, what is the magnitude of the electric field at p

The electric field of a very large (essentially infinitely large) plane of charge is given by:

E = σ/(2ε₀)

E is the electric field, σ is the surface charge density, and ε₀ is the electric constant.

To determine σ:

σ = Q/A

Where Q is the total charge of the sheet and A is the sheet's area. The sheet is a square with a side length d, so A = d²:

σ = Q/d²

Make this substitution in the equation for E:

E = Q/(2ε₀d²)

We see that E is inversely proportional to the square of d:

E ∝ 1/d²

The electric field at P has some magnitude E. Now we double the side length of the sheet while keeping the same amount of charge Q distributed over the sheet. By the relationship of E with d, the electric field at P must now have a quarter of its original magnitude:

c) power lifting, will be your answer.

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