Coulomb's Law Question with Charges on x-axis

by Sabellic
Tags: charges, coulomb, xaxis
Sabellic is offline
Dec25-07, 12:41 AM
P: 68
1. The problem statement, all variables and given/known data

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.

2. Relevant equations
Coulomb's Law:
Fe= k * qq' / r^2

3. 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.
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tongpu is offline
Dec25-07, 12:48 AM
P: 22
For the magnitude it is the absolute value but the direction (negative or positive depends on your axis). Draw a free body diagram..
Oerg is offline
Dec25-07, 06:43 AM
P: 363
Between the +2.0 and the -3.0 charge, the force acts towards the left because of an attraction whereas between the -3.0 charge and the -5.0 charge, the force acts towards the left too because of repulsion. Therefore you should draw a free body diagram to see how the forces of attraction and repulsion add up.

In considering the electrostatic force of attraction, you cannot consider the directional values of the force and the polarity of the charges. You can only start to consider the polarity of charges when considering electric field strength, in which it is a vector and the polarity will tell you the direction of the electric field strength.

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