SOLUTION: find the force acting on a + 10 MuC charge placed 60 cm east of a +6 MuC charge and 40 cm north of a -8 MuC charge (MuC=microcoulomb =10^-6C)

Question 1134598: find the force acting on a + 10 MuC charge placed 60 cm east of a +6 MuC charge and 40 cm north of a -8 MuC charge (MuC=microcoulomb =10^-6C)
Answer by rothauserc(4718) (Show Source): You can put this solution on YOUR website!
The Coulomb Constant(k) is 8.988 10^9 N * m^2/C^2, where N is newtons and m^^2/C^2 is meters squared per Coulomb squared
:
let +10 MuC charge be q1, +6 MuC charge be q2, and -8 MuC charge be q3
:
let F1 be the force acting on q1, F12 be the force on q1 due to q2, F13 be the force on q1 due to q3, then
:
F1 = F12 + F13
:
the above formula seperates into x and y components
:
F1(x) = F12(x) + F13(x)
:
F1(y) = F12(y) + F13(y)
:
F1(total) = F1(x) + F1(y)
:
Note r12 = 60 cm is 0.6 m, r13 = 40 cm is 0.4 m
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Note The point charges q1, q2, q3 are located at the corners of a right triangle
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Setup a rectangular coordinate system with q1 as the origin
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Note F12 is a repulsive force acting along the positive portion of the x axis and F13 is an attractive force acting along the negative portion of the y axis
:
F1(x) = (k * q1 * q2)/r12^2 + 0(no x component for F13(x))
:
F1(x) = (8.988 10^9 * 10^-5 * 6 10^-6)/(0.6)^2 = 149.8 10^-2 = 1.5
:
F1(y) = 0 + (8.988 10^9 * 10^-5 * 8 10^-6)/(0.4)^2 = 449.4 10^-2 = 4.5
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Note F1(y) is negative
:
******************************
F1(total) = 1.5 - 4.5 = -3.0 N
******************************
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