thin aluminum foil

[logistic_guy]

A flat square sheet of thin aluminum foil, \(\displaystyle 25 \ \text{cm}\) on a side, carries a uniformly distributed \(\displaystyle 35 \ \text{nC}\) charge. What, approximately, is the electric field \(\displaystyle \bold{(a)} \ 1.0 \ \text{cm}\) above the sheet and \(\displaystyle \bold{(b)} \ 20 \ \text{m}\) above the sheet?

 

[khansaheb]

A flat square sheet of thin aluminum foil, \(\displaystyle 25 \ \text{cm}\) on a side, carries a uniformly distributed \(\displaystyle 35 \ \text{nC}\) charge. What, approximately, is the electric field \(\displaystyle \bold{(a)} \ 1.0 \ \text{cm}\) above the sheet and \(\displaystyle \bold{(b)} \ 20 \ \text{m}\) above the sheet?

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[logistic_guy]

\(\displaystyle \bold{(a)}\)

We can treat the sheet as infinite sheet as the distance is much smaller compared to the dimension of the sheet.

The electric field for infinite sheet is:

\(\displaystyle E = \frac{\sigma}{2\epsilon_0} \ \ \ \ \) (I will derive this result in future Episodes.)

Then,

\(\displaystyle E = \frac{\sigma}{2\epsilon_0} = 2\pi k\frac{Q}{A} = 2\pi(9 \times 10^9)\frac{35 \times 10^{-9}}{0.25^2} = \textcolor{blue}{31667.3 \ \text{N/C}}\)

 

[logistic_guy]

\(\displaystyle \bold{(b)}\)

The point is considered very far (compared to the dimensions of the sheet), we can treat the sheet as a point of charge.

Then,

\(\displaystyle E = k\frac{Q}{r^2} = 9 \times 10^9 \frac{35 \times 10^{-9}}{20^2} = \textcolor{blue}{0.7875 \ \text{N/C}}\)