Since the field strength decreases with distance and

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Since the field strength decreases with distance, and we are looking for the point that the two individual fieldscancel (have equal magnitude but opposite direction), we must look at a point that is closer to the weaker chargeso that the effect of closer proximity compensates for the effect of weaker magnitude.Scaling SolutionElectric fields scale linearly with charge and inversely with the square-of-the-distance.The left charge is 4xweaker than the right charge, so to compensate, the bead will need to be square-root-of-4 (=2) times closer tothe left charge than the right one.The total distance between the beads is 6.0 meters; so (4m,2m) has the proper2:1 ratio in separations, and also sums up to 6 meters.The bead must be closer to the left charge, the separationto the left charge must be 2m, and the separation from the right charge must be 4m.This puts the zero-field-point at x = -1 cm.Math solution:|E1|= ke*q1/(-3-x)2|E2|= ke*q2/(+3-x)2set |E1| = |E2| ; the kecancelsSo:q1/(-3-x)2= q2/(+3-x)cross multiplying gives :q1 *(+3-x)2= q2*(-3-x)take the square root of both sides : sqrt(q1)*(+3-x)=sqrt(q2)*(-3-x)divide both sides by sqrt(q1) :(+3-x) = [sqrt(q2)/sqrt(q1)]*(-3-x)(+3-x) = sqrt(q2/q1)*(-3-x)(+3-x) = -3sqrt(q2/q1) - sqrt(q2/q1)xsubtract 3 from both sides :-x = (-3sqrt(q2/q1) -3 ) - sqrt(q2/q1)xcollect the x-terms on LHS:-x + sqrt(q2/q1)x= (-3sqrt(q2/q1) - 3)pull -3 out front on RHS :-x + sqrt(q2/q1)x= -3(sqrt(q2/q1) + 1)pull x out front on LHS :x [-1 + sqrt(q2/q1)]=-3(sqrt(q2/q1) + 1)divide both sides by [...] :x=-3(sqrt(q2/q1) + 1) / [ sqrt(q2/q1) - 1]note that:sqrt(q2/q1) = sqrt(16/4) = sqrt (4) =using +2:x=-3 (2+1) / (2-1)This solution is non-sensical because it is not between the two negative chargesusing -2:x=-3 (-2+1) / (-2-1)=-3 (-1) / (-3)so x = -1m is the valid solution.Alternative Math Solution :Solve [q1 *(+3-x)2= q2*(-3-x)2] for x using the quadratic equation solution.22±2=-9=-1
7.Which of the following scenarios is physically possible?(i) An object of mass 4.0 μg and net charge of+6.4 x 10-19Cis split into two smaller objects of masses 3.0 μgand 1.0 μg with net charges of+4.8 x 10-19C and +1.6 x 10-19C, respectively.(ii) Three identical objects, each with the same mass and each carrying a single elementary charge, arecombined to form an object with three times the individual masses and zero net charge.(iii) Two identical objects, each with a mass of 3.0 μg and each carrying a net charge of-0.8 x 10-19C, arecombined into an object of mass 6.0 μg and carrying a net charge of and -1.6 x 10-19CSolution

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Term
Winter
Professor
NoProfessor
Tags
Electric charge, charges Q