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Old Jan 16th 2009, 06:53 PM   #1
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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.
Aryth is offline   Reply With Quote
Old Jan 19th 2009, 03:25 AM   #2
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Force can not be added as such .they need to be addded vectorially.so the solution may be wrong.
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