Net force on a particle.... Consider the following two \" fixed\" charges with l
ID: 1505840 • Letter: N
Question
Net force on a particle....
Consider the following two " fixed" charges with locations in a plain given by x and v (x., Y, q): a: (-1.00cm, 2.00cm, -5,0 mu C) b: (3.00 cm, -1.00 cm, + 3,0 mu C) Find the magnitude of the force on b from alpha. and then give the force in component form (i.e. x-component and y-component).Now assume this region has an additional uniform electric field (from some external source), with a magnitude of 1.5x10^ ' V/m directed 60^ degree above the + x axis. find the force on particle Beta from this external field and give it in component form Combine the two forces on Beta (from particle and from the external field) via vector addition Give the net force in component form.Explanation / Answer
distance between a and b = sqroot((3+1)^2 + (2 + 1)^2) = 5
1. Force on b due to a = k(-5)(3)*10^-12 / 0.05^2 = -53.34 N
In component form, F = 53.34(4/5 - 3/5 ) = 42.672 i - 32.004 j
2. Force due to field = 1.5*10^7*(-5*10^-6)(cos(60)i + sin(60)j)
-37.5 i - 64.952 j
3. Net F = -37.5i + 42.672i - 64.952j - 32.004 j
F = 5.172i - 96.956j
4. |F| = sqroot(5.172^2 + 96.956^2) = 97.0938 N
Direction = arctan(-96.956/5.172) = -86.946 deg with x axis