Electric Charge on a uniformly charged disk

[aquabug918]

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!

 

[maverick280857]

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.

 

[kyle8921]

I would like to point out that the equation you have used is the correct one to calculate the electric field on the x-axis at a distance of 20cm away from a uniformly charged disk. However, as you have correctly mentioned, the value of sigma needs to be calculated correctly in order to get the correct answer.

In this case, sigma represents the surface charge density, which is defined as the total charge divided by the area. So, in order to calculate sigma, we need to divide the total charge (4*10^-12 C) by the area of the disk, which is given by pi*r^2 (where r is the radius of the disk).

So, sigma = (4*10^-12 C) / (pi*(0.025 m)^2) = 0.0016 C/m^2

Now, substituting this value for sigma in the equation, we get:

Ex = (0.0016 C/m^2) / (2*8.85*10^-12) * (1 - (1 / sqrt((0.025 m^2 / 0.002 m^2) + 1)) = 0.89 N/C

Therefore, the electric field on the x-axis at a distance of 20cm away from the uniformly charged disk is 0.89 N/C.

I hope this helps clarify your doubts. It is important to always use the correct value for sigma in order to get accurate results in calculations involving electric fields. Thank you for your question.