Question

Problem #3 1/2 ROZW Two insulated rods of length L = 1 m have a linear charge density of 1 = 1 nc/m. a- What is the electric field at P due to the vertical rod? b- What is the electric field at P due to the horizontal rod? C- What is the total electric field at P? Some charges are removed from the horizontal rod so the charge density is now - 2. d- What is the new total electric field at P? We recall the elementary field for a vertical rod due to a dq charge: 1/2 1 4/2 1 42 dĒ = (cos8 + siney) 22 de 47TED

Answer 1

Idar O; TCA .. the electrie At-- o "As we know the elementary electowe field will be d = 1 (caso at sino ŷ) 2 / do Nows fort Bo vertical rod, all the components of the eleementary electric field along y-direction will be cancelled out by the opposite elements of the rod on the 4 both side of the point o field due to the Vertical rod at point p will be a a-direction. (positive n-direction) Of .: the deetuje field, Eu the scoso do o; and of are initial and final angle made by the point Pat A & B. at A, using A ACP, tan bis O et =Dia tano (1) - similarly at By Of A but we take 0; -- because we have to go cloekwise direetion from PO, 2. the electric field Eva 22 l caso do â are med om The Simola we have into (sing tsin py) * • , * * * , - da dne/m= 109 c/m. Lim. A = 9x103 kg. 1354 Fi - 5x10° y 2482x 10" /m är Ēr ? 25.455 vmâ 25.46 V/m Sil i

" As we know the elementary electric field will be d = 1 (Sino a + Cose }) 212 do Ако - 27 22 ✓ Now, for horizontal rod all the components of the eleqmentary electric field along direction will be cancelled out by the opposite elements of the rod on the 4 Toi both side of the point O2 the a electrie field due to the B horizontal rod at points will be along y-direction. (positive direction) or :: the deetrie field, Eki teha ? Cosodo ĝ O; and of are initial and final angle made by the point Pat B & c. at B, using a Bop, a tan biz de Bones • Biz tari" (1). - similarly at e, Of a but we take 0; -w because we have to go cloekwise direction from POL- ... the electric field Egg ano 21 l coso do ý tra we anto (*sing +Sin 7) ģ The 9txã às PSI have da dne/m= 109 c/m. La ime A = 9 x 100 kg. mist s Ēv - 3x10° x 2592 x 10" w/ my Ev. 25.455 vmâ -25.46 ym

due to both vertical and c) the net electric horizontal rod field at e is. On1 292da AñE 2. Vim. and y AC - 2 2 2 2 + ane 22 tý a) tate 2524 ) /m. = 25,46 å +25469 lĒT E = magnitude of the electrie field is C = The 2.120 x J je + ja yom cart. 44 Von - 3x10° x 4x100m Ylm = 36 v/m - d if no for vertical d changes to do the horizontal electric field will beco(due to the horizontal rod) - 1 252 ATTEOT so N E -- The 25EA ĝ .. it will be acting along negative y direction, and the total electric fidd will be- Ē - Fy + Fit Fate OSA A - The sea ģ - 27 - -4) v/m. magnitude will be- E 1 252 x (1²+12)² Yom /m. = gx10° x 4X10 Vlm = 36 V/m. ie the magnetude will be same but the direction will change to ao from previous,

due to both vertical and c) the net electric horizontal rod field at e is. On1 292da AñE 2. Vim. and y AC - 2 2 2 2 + ane 22 tý a) tate 2524 ) /m. = 25,46 å +25469 lĒT E = magnitude of the electrie field is C = The 2.120 x J je + ja yom cart. 44 Von - 3x10° x 4x100m Ylm = 36 v/m - d if no for vertical d changes to do the horizontal electric field will beco(due to the horizontal rod) - 1 252 ATTEOT so N E -- The 25EA ĝ .. it will be acting along negative y direction, and the total electric fidd will be- Ē - Fy + Fit Fate OSA A - The sea ģ - 27 - -4) v/m. magnitude will be- E 1 252 x (1²+12)² Yom /m. = gx10° x 4X10 Vlm = 36 V/m. ie the magnetude will be same but the direction will change to ao from previous,