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# phys2_exm1 - Coulombs Law F=(1 4pi0(q1*q2 r2(rhat F(newtons...

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Coulomb’s Law F=(1 / 4piε 0 )(q1*q2 / r 2 ) (r hat ) F (newtons) attractive if q’s opp signs 1 / 4piε 0 = 9*10^9 Nm^2 / C^2 F=qE (F&E vectors) E lines from + to – Scalar V= - ∫(E∙ds) Pt chg: E= (q / 4piε 0 r 2 ) (r hat ) V/M; V= - ∫(E∙ds) = (q / 4piε 0 r) Wk to move chg: W = ∫(F∙ds) = ΔU = Δpot energy = qΔ(V) Gauss’s Law ∫closed area(E∙dA)=q encl 0 Relates chg INSIDE closed surface to flux of elec fld coming through surface Φ E =flux of E=q/ε 0 = ∫clsdar(E∙dA); dA=an AREA increment Derive Coulomb’s from Gauss’s: pt Q ∫clsdar(E∙da)= q/ε 0 , E=q/(4pi ε 0 r 2 ) * r hat Place test chg on Gaussian suface F=qE=(q*q0)/(4pi ε

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phys2_exm1 - Coulombs Law F=(1 4pi0(q1*q2 r2(rhat F(newtons...

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