theses were given on a test that I missed and I still dontunderstand how to gert
ID: 1748914 • Letter: T
Question
theses were given on a test that I missed and I still dontunderstand how to gert the correct answer 1) Two charges, +q and +2q are fixedat a separation of1m along the x-axis. Find the distance from the +q charge at whichthe electric field is zero. express your anser in meters.... thecorrect answer is .414 ...but how do you get that 2) a 1e-9 C charge is placed one meter from a 4e-9 C cahrge.If I place a 2E-9 charge 1/3 m from the 1e-9C what force does itexperience due to the first two charges? answer in newtons.....thecorrect answer is 0..but how do you get it?? theses were given on a test that I missed and I still dontunderstand how to gert the correct answer 1) Two charges, +q and +2q are fixedat a separation of1m along the x-axis. Find the distance from the +q charge at whichthe electric field is zero. express your anser in meters.... thecorrect answer is .414 ...but how do you get that 2) a 1e-9 C charge is placed one meter from a 4e-9 C cahrge.If I place a 2E-9 charge 1/3 m from the 1e-9C what force does itexperience due to the first two charges? answer in newtons.....thecorrect answer is 0..but how do you get it??Explanation / Answer
1) if a charge is placed between the +q and the +2qcharges it will pulled a different direction by each charge. In order for the electrical field to be 0 the field from eachcharge must be the same.Ke is a constant. r is the distance from the +q charge. Sincewe are discussing a point between the two charges and thetotal distance between the two charges is 1m the distance to the+2q charge = 1m-r Since we know the Electric field = Ke*|q|/r^2, and the fields have to be the same we cansay: Ke*|+q|/r^2 = Ke*|+2q|/(1-r)^2 then we can cancel out the Ke's and the|+q|'s to get: 1/r^2 = 2/(1-r)^2 we can cross multiply to get: 1*(1-r)^2 = 2*r^2 then simplify to get: 1-2r+r^2 = 2r^2 subtract 2r^2 from both sides: 1-2r-r^2=0 substitute numbers into the quadratic equation - in case itbeen awhile the equation is(-b±(b^2-4ac))/(2a). From the equation above a= 1, b = -2, c = -1. This gives us: (--2±((-2)^2-4(1)(-1)))/(2(1)) =(2±(4+4))/2) = (2±8)/2 ={-.414,2.414} since the distance is not more than 1m, it must be .414m (thenegative sign doesn't matter, it just depends on which distance youchose to be r) 2) Coulomb's Law states: force = Ke*|q1|*|q2|/r^2 (where Ke is Coulomb's constant, equalto 8.9875 * 10^9 N*m^2/C^2, and ql andq2 are the charges of the two particles,and r is the distance seperating the two particles) so for the force on the 1e-9 C charge from the 4e-9 C chargeis equal to: F = 8.9875 * 10^9 N*m^2/C^2 * |1 * 10^-9 C| * |4 * 10^-9 C|/(1m)^2 = 3.595 * 10^8 N Since the 2e-9 C charge is placed 1/3 m from the 4e-9 C,it must be 1 m - 1/3 m from the 1e-9 C charge, so r = .7 m and F =8.9875 * 10^9 N*m^2/C^2 * |1 * 10^-9 C| * |2 * 10^-9 C|/(.7 m)^2 =3.595 * 10^8 N So these forces would cancel out and thus the answer is 0(although I would've thought they would have to be of oppositesign, are you sure the sign on both the charges ispositive?) Sorry this explanation got so long, I hope it is clear andhelpful.
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