Question
Find the electric field at a point P due to a 10 × 10^−9C charged disk of radius 2cm. The charges are fixed and uniformly distributed on the disk. The point P is 33 cm away from the disk. Now you drill a hole of radius 0.35cm at the center of the disk. Find the electric field of this new ”disk” without taking the difference between the two disks, rather solve by integration from the small to bigger radius.

It is well known that the electric field at a perpendicular distance z from the origin of a circular wire with charge Q and radius r is given by  Hence the electric field at a perpendicular distance z from the origin of a circular disk with charge Q and radius R is given by  where dQ  is the charge of the elementary ring of radius r and thickness dr.

Now, in this case, , where is the surface charge density of the disk , i.e., .

Hence .

Here , , and .

Hence After drilling the hole of radius R0=0.35 cm, the electric field becomes since in this case, the area of the disk is Hence #### Earn Coins

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