Question

A flat square sheet of thin aluminum foil, 25 cm on a side, carries a uniformly distributed+42 nC charge.What, approximately, is the electric field at the following positions?

a. 1.0 cm above the center of the sheet.
b. 20 m above the center of the sheet

Answer 1

To solve this problem we will apply the concepts related to the surface charge density. This value will be found between the load and the Area. With this value it will be possible to find the value of the electric field, since the latter depends on the surface charge density and the permittivity of free space.

a) The surface charge density is

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

$\sigma = \frac{42*10^{-9}C}{(0.25m)^2}$

$\sigma = 672*10^{-9} C/m^2$

A. 1.0 cm above the center of the sheet,

$E = \frac{\sigma}{2\epsilon_0}$

Here,

$\epsilon_0$ = permittivity of free space

Replacing,

$E= \frac{672*10^{-9}}{2 (8.85*10^{-12})}$

$E=37966N/C$

Therefore the electric field at 1cm above the center of the sheet is 37966N/C

b) For 20m, the distance becomes considerably large so it is necessary to take the sheet as a point load

$E= \frac{kQ}{r^2 }$

Here,

k = Coulomb's constant

Q = Charge

r = Distance

$E= \frac{(9*10^9 ) (42*10^{-9})}{(20)^2}$

$E=0.945N/C$

Therefore the electric field at 20m above the center of the sheet is 0.945N/C

Answer 2

(a) The electric field at 1 cm is 37966 N/C

(b) The electric field at 20 m is 0.945 N/C

Electric field:

(a) The distance of 1 cm is very near to the sheet as compared to the size of the sheet. The electric field due to a uniformly charged sheet is given by:

$E = \frac{\sigma}{2\epsilon_o}$

where σ is the charge density which is given by the total charge upon the area of the sheet,

$\sigma=\frac{q}{A} = \frac{42\times10^{-9}}{0.25\times0.25}\\\\\sigma=6.72\times10^{-7}C/m^2$

So, electric field:

$E=\frac{6.72\times10^{-7}}{2\times8.85\times10^{-12}} \\\\E=37966N/C$

(b) The distance of 20m is very far from the sheet as compared to the size of the sheet. So we consider the sheet as a point charge. The electric field due a point charge is given by:

$E=\frac{1}{4\pi \epsilon_o} \frac{q}{r^2}$

where r is the distance = 20m

q is the charge = 42 nC

$E=9\times10^9\times\frac{42\times19^{-9}}{20\times20}\\\\E= 0.945\;N/C$

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