- #1

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The Electric field is, after some calculation, is given by E_p below:

z is the z-axis, and a is the radius of the disc.

Now for the questions at the bottom of the page, here are my thoughts:

σ is independent of a because as a->2a, Q->4Q, and a^2 -> 4a^2. Same for all changes of scale.

As a -> infinity, E_p -> σ/(2ε_0)

But the second one I'm not sure about.

As a -> 0, surely the charge density stays constant i.e. at σ. Thus the field appears to be the 0 vector. However, we know that the field due to a point charge is proportional to

**r^**/r^2, where

**r^**is the outward/inward unit vector and r is it's magnitude.

I know our area tends to 0 as a -> 0, so our charge Q must tend to 0 also. But this means our field from the "point charge" must be 0, disagreeing with the usual electric field for a point charge...