Question

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

Answer 1

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:

$E_{new} = E/4$

Answer 2

When the sheet is stretched so that its sides have length 2d, the magnitude of the electric field at P will be    $\dfrac{E}{4}$

What will be the magnitude of the electric field at P?

To calculate the electric field intensity on a large plane shape will be given by

$E=\dfrac{\sigma}{2\varepsilon _0}$

Here:

E= is the electric field

σ= is the surface charge density

ε₀ =is the electric constant.

Now the surface charge density is given by

$\sigma =\dfrac{Q}{A}$

Here:

Q =is the total charge of the sheet

A= is the sheet's area.

Since the elastic sheet having length = d

$A=d^2$

So:

$\sigma=\dfrac{Q}{d^2}$

Putting the value of   $\sigma=\dfrac{Q}{d^2}$    for in the electric field equation  

$E=\dfrac{Q}{2\varepsilon _0d^2}$

Since the value E is inversely proportional to the square of the length of a sheet

Now if we put  $d'=2d$

$E'=\dfrac{Q}{2\varepsilon _0 (2d^2)}=\dfrac{Q}{2\varepsilon _0 (4d^2)} =\dfrac{\sigma}{2\varepsilon_0 4} =\dfrac{E}{4}$  

Thus When the sheet is stretched so that its sides have length 2d, the magnitude of the electric field at P will be    $\dfrac{E}{4}$

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