"Two charges, q(adsbygoogle = window.adsbygoogle || []).push({}); _{1}, and q_{2}, are located at the origin and at (0.50m, 0), respectively. Where on the x axis must a third charge, q_{3}, of arbitrary sign be placed to be in electrostatic equilibrium if (a) q_{1}and q_{2}are like charges of equal magnitude, (b) q_{1}and q_{2}are unlike charges of equal magnitude, and (c) q_{1/sub] = +3.0 microCoulombs and q2 = -7.0 microCoulombs?" for (a), I thought since they are like charges, q3 can only be in electrostatic equilibrium if it's between the two charges. let r = distance from (0,0) F31 = F32 (kq3q1)/r2 = (kq3q1)/(0.50-r)2 r2 = (0.50-r)2 r = 0.25 m but the answer is -0.25m. Is that a typo? for (b), I thought since they are opposite charges, q3 can only be in equil. if it's not between the two charges. let r = distance from (0,0) or (0.50, 0) F31 =- F32 (is it correct to put the negative sign here?) (kq3q1)/r2 = -(kq3q1)/(0.50+r)2 r2 = (0.50+r)2 0 = 0.50 but this doesn't make sense, so there is no location for q3 to be in equilibrium. (This answer is correct, but I think I could've solved it by just thinking about it... but I'm not sure how) for (c), I thought the only place for q3 (regardless of its sign) is to the left of (0,0) F31 =- F32 (kq3q1)/r2 = -(kq3q1)/(r+0.50)2 q1(r+0.50)2 = -q2r2 (q1 + q2)r2 + q1r +0.25q1 = 0 I use the quadratic formula, and take only the negative root, since r < 0 r = -0.20m. So q3 would be at (-0.20,0). but the answer here is -0.94m}

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# Electrostatic equilibrium

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