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Homework Statement
Three point charges are placed at the following points on the x-axis: +2.0 micro Coulombs at x=0; -3.0 micro Coulombs at x=40 cm; and -5.0 micro Coulombs at 120 cm. Find the force on the -3.0 micro Coulomb charge.
Homework Equations
Coulomb's Law:
Fe= k * qq' / r^2
The Attempt at a Solution
For 2.0 microC charge to the -3.0 microC charge:
Fe= (9.0 * 10^9) * (2.0 * 10^-6)(-3.0 * 10^-6) / 0.40^2
= -54 * 10^-3 / 1.6 * 10^-1
= -0.3375 N
For -5.0 microC charge to the -3.0 microC charge:
Fe= (9.0 * 10^9) * (-5.0 * 10^-6)(-3.0 * 10^-6) / 0.80^2
= 135 * 10^-3 / 6.4 * 10^-1
= 0.2109 N
The book says the answer is -0.55 N. But I thought that with this procedure we are supposed to use the absolute values. I don't understand which Fe I should be subtracting from the other. Please help me.