Question

strie field associated with charge distributed evenly ell located at its center. Furthermore, the 1. The Shl There is a statement about the electric field associated with charge din over a spherical shell. The theorem states that at all lehall The theorem states that at all points outside the shell the electric field as that of a main charge containing all the charge on the shell located at its center the them states that the electric field inside the shell is ZO. In this problem, we will prove both this problem, we will prove both parts of the theorem by directly computing the electric field in and outside the shell. The electric field at a point on the 2-axis corresponding to a unitatily charged circular ring lving in the ry-plane was derived in class. For a ring of radius R carrying total charge Q the field is. AQ2 E.(2) (R + 223/2 Aspherical shell can be decomposed into a stack of rings of varying radius, cach containing an infinites imal amount of charge dQ. Each ring in this stack contributes to the net electric field an amount de The form of dE is that as above with Q replaced with d and R and x modified to match the radius and distance to each ring. Below a contributing ring is shown for a sphere of radius R containing a charge Q. The infinitesimal ring carrying charge dQ is located at and has radius r. The ring contributes dE, to the electric field at . Also in the diagram is 8, the sphere's polar angle, do, the angular width of the the ring, and the distance from the field point at 2 to any point on the ring. DE dQ

Answer 1

Given that: - A spherical shell with radius R containing charge .. we have to find out the electric field on Za axig at distance z from the centere of sphenico cell by considering that spherical cell meede by stacking ring havinge radios y and centers on the z-axis where y'is variable from o to R. - S Hz dQ: charge here the ring the triangle opQ 9 from figure consider By using low of cosine se = ² + 2 = 2 R2 Caso

b). From the . s² = here s, and a constant, aking partial R +3² Q RZ Cego core rasiable and - R and a con differential of equation o esds - oto- ERZ (-simo da) lsd = 2 R2 sind do sds = Rz simodo - dEz = - ① kada R2 22 7 % to find da wsing the concept of surface charge dinsity which is constant for uniform charge distribution. I da= aar. Rolo dң (т) (gas) чле- sogla da RdO dQ = a z do

so dby K2 693229312 ( 2 (7-2'53) do I = smo » r= Rsimo dQ = 2 R R$mo do = Simone de obtamed in pcent ☺. using the scalation R2 SM@ de sds = do= so sds Rasino s ds F 282 qs put it in equation ④ do Sime. sds 2 Resmo putting this relation in equation ③ we have - KS as = = ds RZ ns d'Ez = RM. - x (R+z2% 2(R²+223/2 - spherre: ce) Electric outside the field s varies from RR to 2+R

24R 2+R Reds 1 20422) / 2-R 2-R 212+R K 2 (R² + 2 2 3 3/2 se (tzeaf(225€] See the most am 222-4*R*4.27R] Enout = RM2R Y 12² +22) 72 at a long distance, for 2» oudside R²2 a 22 1 So Е - кR 72 К2К , 9,3 Ezout 127 Езот и cell is Ē is imursly behaviour of electric field outside the equivalent to the point charge as proportional to the distance.