Posted by **Nat** on Thursday, February 22, 2007 at 3:54pm.

1)Three positive particles of charges 9µC are located at the corners of an equilateral triangle of side 15cm. Calculate the magnitude and direction of the force on each charge.

2)What is the electric field strength at a point in space where a proton (m = 1.67 x 10-27 kg) experiences an acceleration of 1 million "g's"?

I have some problem setting up these problems. Could you please help me?

1. The force on each particle is the resultant of the two force due to the other two particles. Use Coulomb's law

K - k e^2/r^2

to compute the force between two particle of charge e that are a distance r apart. Because of symmetry, only the force component perpendicular to the line between the other two particles will be nonzero.

2. E = F/e = a/(m e)

a = 9.8*10^6 m/s^2 (one million g's)

e = 1.6*10^-19 Coulombs

m = m = 1.67 x 10-27 kg

The units of the field strength E will be Newtons/Coulomb which is the same as Volts/meter

For (1), I meant to write

F = k q^2/r^2, where

q = 9*10^-6 C, r = 0.15 m, and k is the Coulomb's-law constant, which you should look up. That is the force between two particles. For the resultant force acting upon one particle due to the two others, imagine one particle placed along the y axis and the other two along the x axis. The resultant force on the particle on the y axis will be in the y direction. You only have to add the two components of force along the y axis. The other components cancel out. The component along the y axis is F cos 30

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