Electric Charge on a uniformly charged disk

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To find the electric field on the x-axis from a uniformly charged disk, the surface charge density (σ) is calculated as the total charge divided by the area of the disk. For a disk with a radius of 2.5 cm and a total charge of 4*10^-12 C, σ is determined using the formula σ = Q/A. The electric field (Ex) can then be calculated using the provided equation, which simplifies to Ex = (σ / (2*ε₀)) * (1 - (1 / √((R²/x²) + 1))). The final calculated electric field at a distance of 20 cm is 0.89 N/C.
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Last question I promise,,,,

A uniformly charged disk has a radius of 2.5 cm and carries a total charge of 4*10^-12C.

Find the electric field on the x-axis at a distance of 20cm away.

I used the equation:

Ex = (sigma/ (2*Eo)) * (1 - ( 1/ sqrt( ( R^2/x^2) + 1))

this is what i did so far

Ex = (sigma/ (2*8.85*10^-12)*(1 - ( 1/ sqrt( ( 0.025m^2/0.002m^2)-1)

I am not sure how to calculate sigma in this case. I am guessing that it is the total charge dived by the area. But, i think i am missing something.

The answer is 0.89 N/C

Thanks!
 
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No you are not missing anything. \sigma is indeed the surface charge density and is equal to the total charge divided by the total area.

You could have verified this by dimensional analysis too. The expression really isn't as complicated as it seems.
 
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