Physics Help Forum Electrostatic Force

 Jan 16th 2009, 06:53 PM #1 Junior Member   Join Date: Apr 2008 Posts: 21 Electrostatic Force Four particles form a square. The charges are $\displaystyle q_1 = q_4 = Q$ and $\displaystyle q_2 = q_3 = q$. Just for your info, the particles are arranged as such: 1 2 3 4 (a)What is $\displaystyle \frac Qq$ if the net electrostatic force on particles 1 and 4 is zero? Here's my solution for part a: (My notation for Force for particles is this, say I want to say the force of particle 2 on 1, I write $\displaystyle F_{21}$) $\displaystyle \sum F_1 = F_{21} + F_{31} + F_{41}$ This says that the net force on 1 is the sum of the forces on 1 from 4, 3, and 2. $\displaystyle \sum F_1 = k\frac{qQ}{a^2} + k\frac{qQ}{a^2} + k\frac{Q^2}{(\sqrt{2}a)^2}$ $\displaystyle = 2k\frac{qQ}{a^2} + k\frac{Q^2}{2a^2}$ $\displaystyle 2k\frac{2qQ + Q^2}{2a^2} = k\frac{2qQ + Q^2}{a^2}$ Turns out I get the same thing for $\displaystyle \sum F_4$ because the net force for Q is 0. So, now I get the net force on Q: $\displaystyle \sum F_Q = 2k\frac{2qQ + Q^2}{a^2} = 0$ $\displaystyle \frac{2k}{a^2}(2qQ + Q^2) = 0$ Since k and a are obviously not 0, then $\displaystyle (2qQ + Q^2)$ must be 0, so: $\displaystyle 2qQ + Q^2 = 0$ $\displaystyle Q^2 = -2qQ$ $\displaystyle Q = -2q$ $\displaystyle \frac Qq = -2.0$ Is that right? If so, here's part b. (b) Is there any value of q that makes the net electrostatic force on each of the four particles 0? If my answers aren't right, could you please correct them and explain so I can better understand how to do other problems of this nature. Thanks.