Download Electric Fields from Multiple Charges and Charge Distributions and more Slides Electromagnetism and Electromagnetic Fields Theory in PDF only on Docsity! Multiple Charges and Charge Distributions Docsity.com • Assume a large number (essentially continuous distribution) of charges. – The contribution to E-field at some point (x,y,z) from dV: + + + + + + + + + + + ),,( zyxP r̂ r r ˆ ˆ4 1 2 dvEd o ρ πε = Docsity.com • In general, – ρ can vary throughout the integration volume – r2 varies throughout the integration volume. – the direction of the r can vary throughout the integration volume. vector. unit dependence 2r / ondistributi charge of spaceover integral factor P point, at fieldE ××= − ∫ Docsity.com • Determine the Coulomb force and E-field at (0,0,4) due to four charges of 10 µC at: – (-3,0,0), – (3,0,0), – (0,-3,0), and – (0,3,0). • q = 20 µC at (0,0,4) Docsity.com y x ∑ − − = i i i o rr rrQE 3ˆˆ ˆˆ 4πε EF q= Docsity.com • The E-field vector at (0,0,4) points upwards in the z-direction. Docsity.com • The force on the charge, q = 20 µC at point (0,0,4) is: ( ) Nz m kVCEqF ˆ234.0 52.1120)4,0,0( = == µ Docsity.com • Electric Field of a Ring of Charge. – Ring of positive charges in free space. – Choose coordinate system, so that z-axis is at the center of the ring. • What is E at source point on the z-axis, P(0,0,h)? + + + + + + + + + + b dφ dl=bdφ Docsity.com • is independent of φ. • depends on φ. – but, opposite components cancel! φ π πε ρ bd hb hb o l ∫ + +− = 2 0 2 3 22 ˆˆ 4 zrE ẑ r̂ + + + + + + + + + + b ' rr EE −= rE ' rE Docsity.com • Integration is simplified! ∫ + = π φ πε ρ 2 02 3 22 ˆ 4 d hb h o lb zE π2 2 3 22 ˆ 4 2 + = hb h o lb zE πε ρπ Docsity.com • Note that: – the total charge which implies: • The field points upward • Its magnitude approaches zero for large h • The greater the radius, b, the lower the field at point P(0,0,h). Qlb =ρπ2 2 3 22 ˆ 4 + = hb h o Q zE πε Docsity.com • As in the previous exercise, the opposite components of cancel. E ∫∫ ++ = π φρ πε 2 00 22222 ˆ 4 1 ddr hr h hr r a s o zE Independent of both and . φ̂r̂Constant Docsity.com ( ) ∫∫ + = π φ πε ρ 2 00 2 3224 ˆ ddr hr rh a o s zE ( )∫ + = a o s dr hr rh 0 2 3224 ˆ2 πε πρ zE Docsity.com • Note: ( ) ( )∫ + − = + 2 1222 322 1 Cx dx Cx x Docsity.com • Since – it follows that zE ˆ1 2 222 + −= ha h a Qh oεπ saQ ρπ 2= Docsity.com • So what? – As expected the field points upwards – Its magnitude approaches zero for large h. – The field is proportional to the total surface charge. – The larger the disk, the lower the field at a given point, P(0,0,h) Docsity.com • How about the E-field for an infinite sheet of charge. – For an infinite sheet of charge: • and ( ) zzE ˆ 2 ˆ11 2 2122lim o s o s a hah h ε ρ ε ρ = + −= ∞→ ∞→a 0 Docsity.com
