# Electric potential of four charges

by Metalsonic75
Tags: charges, electric, potential
 P: 29 There are four spheres arranged as the four corners of a square, each sphere having a charge of 10 nC and seperated from each other by a distance of 1 cm. The four 1.0 g spheres are released simultaneously and allowed to move away from each other. What is the speed of each sphere when they are very far apart? (see attached figure). I know that for kinetic and potential energy deltaK = -deltaU K_f - K_i = U_f - U_i K_i = 0 U_f = 0 K_f = U_i First I found the total potential energy of the system. U = (q_1*q_2*k)/r. Since all charges are the same, the interaction of any two charges is q^2 (where q^2 = 1*10^-16). The six interactions between the charges produce the following calculation for potential energy: U = K { q^2/r + q^2/r + q^2/r +q^2/r + q^2/sqrt(2)r + q^2/sqrt(2)r } U = K { 4q^2/r + 2q^2/sqrt(2)r) } Plugging in 10^-8 for q, and 0.01 for r, gets U = K (5.414*10^-14) U = 0.000487 Since U_i = K_f, 0.000487 = 0.5mv^2 The mass of the spheres is 1g (0.001 kg) 0.973 = v^2 v = 0.987, which should be the final speed of each sphere. This is incorrect, though. Could somebody please tell me where I went wrong? Thank you for your time. Attached Thumbnails
HW Helper
P: 2,250
Hi Metalsonic75,

 Quote by Metalsonic75 Since U_i = K_f, 0.000487 = 0.5mv^2 The mass of the spheres is 1g (0.001 kg) 0.973 = v^2 v = 0.987, which should be the final speed of each sphere. This is incorrect, though. Could somebody please tell me where I went wrong? Thank you for your time.
I can't see your attachment yet, but I think the equation in bold is incorrect; can you see why? Remember that the potential energy term on the left was from all four particles.
 P: 29 QUOTE: Remember that the potential energy term on the left was from all four particles. Ah! Thank you! I now have the correct answer!

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