If one could please break this problem down in pieces I would be grateful A very

Question

If one could please break this problem down in pieces I would be grateful

A very, very large flat sheet of paper carries charge uniformly distributed over its surface; the amount of charge per unit area is sigma (C/m^2). A hole of radius R has been cut out of this paper as shown in the figure. (i) Find the electric field Eon the axis of the hole for all z. (ii) Using your expression derived in (i), take the limit of E as z rightarrow plusorminus infinity. How does it compare with the "hole-less" sheet of charge (i.e., R-0)?

Explanation / Answer

part i.

consider a small patch of area dx*dy at coordinate (x,y)

charge=dq=pho*dx*dy

field at the point (0,0,z):

field direction=(0,0,z)-(x,y,0)

=(-x,-y,z)

=(-r,0,z) in cylindrical coordinate system

distance=sqrt(r^2+z^2)

then unit vector=(-r,0,z)/(r^2+z^2)

field magnitude=k*dq/distance^2

=k*pho*dx*dy/(r^2+z^2)

in vector notation, field =dE=(k*pho*dx*dy/(r^2+z^2)^1.5)*(-r,0,z)

due to symmetry, only z component will remain.

hence dE=k*pho*z*dx*dy/(r^2+z^2)^1.5

in cylindrical coordinate system , area of the small segment=r*dr*dt

where t is angle .

then dE=k*pho*r*z*dr*dt/(r^2+z^2)^1.5

then total electric field=integration of dE

=(k*pho*z)*(integration of r*dr/(r^2+z^2)^1.5)*(integration of dt)

here limit of r is from R to infinity, and t is from o to 2*pi

E=(k*pho*z)*(1/sqrt(R^2+z^2))*2*pi

part b:

as z tends to infinity, sqrt(R^2+z^2)=sqrt(z^2)=z

hence E=k*pho*z*(1/z)*2*pi

=k*pho*2*pi

which is same as electric field due to hole-less sheet.

note: here k=coloumb;s constant=1/(4*pi*epsilon)

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