A charge 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 2d, what is the magnitude of the electric field at P?

E/4

Explanation:

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

E = σ/(2ε₀)

Where;

E is the electric field

σ is the surface charge density

ε₀ is the electric constant.

Formula to calculate σ is;

σ = Q/A

Where;

Q is the total charge of the sheet

A is the sheet's area.

We are told the elastic sheet is a square with a side length as d, thus ;

A = d²

So;

σ = Q/d²

Putting Q/d² for σ in the electric field equation to obtain;

E = Q/(2ε₀d²)

Now, we can see that E is inversely proportional to the square of d i.e.

E ∝ 1/d²

The electric field at P has some magnitude E. We now double the side length of the sheet to 2L while keeping the same amount of charge Q distributed over the sheet.

From the relationship of E with d, the magnitude of electric field at P will now have a quarter of its original magnitude which is;

E_new = E/4

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if they have a finite resultant