Coulomb's Law Question with Charges on x-axis


by Sabellic
Tags: charges, coulomb, xaxis
Sabellic
Sabellic is offline
#1
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.
Phys.Org News Partner Science news on Phys.org
Better thermal-imaging lens from waste sulfur
Hackathon team's GoogolPlex gives Siri extra powers
Bright points in Sun's atmosphere mark patterns deep in its interior
tongpu
tongpu is offline
#2
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
Oerg is offline
#3
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.


Register to reply

Related Discussions
estimating charges coulomb's law Advanced Physics Homework 1
estimating charges coulomb's law Introductory Physics Homework 6
Coulomb's Law: forces between charges Introductory Physics Homework 3
coulomb's law and different charges Introductory Physics Homework 2