Question

1 CIRCULAR DISK Com sdke a le r . 1.1 5 POINTS Set up a coordinate system with the centre of the disk at the origin. 1.2 15 POINTS Calculate the electric field a distance z above the centre of the disk. 1.3 10 POINTS How does this compare to the electric field a distance z above the centre of a rod of length 2R with the same total charge?

1.1 1.2 1.3

Answer 1

First, we will calculate electric field due to a ring, the integrating this with appropriate limit we can get, electric field due to disc. Electric field due to charge element on ring: (Consider charge element as rings with linear charge density 2) . → Enor

1 ir do 41€ (r2 + z2) Due to symmetry horizontal field is zero: Enor = 0 1 ar do sin 8 dEyer = 47eg (r2 + z2) JE Z Ar do dEver - 4te(v2 + z2)3/2 Satever Creations So, we got field due to ring is: - Pring rz 2 € (r2 + z2)3/2 Electric field due to disc: For disc with surface charge density o, consider charge element as rings with linear charge density 1:

i rz 2 € (r2 + z2)3/2 - Using linear charge density in terms of surface charge density - 2= ods 2 rdr 2tr2tr -= odr de = c rz dr de 2 €(r2 + z23/2 pro rz dr J. 2 € (r2 + z2)3/2 Solving this integral we get =(-WRT) Charge on disk = onR2 For a rod of length (2R) linear charge density is 1 =

So electric field will be - - az dr [de - Le trebalo de = 3/2 2 411€0(y2 + z273/2 Solving the integral we get - 3-2002 (what) Now you can compare both results.