Electric charge and total force

[Rasine]

Three charges, Q1, Q2, and Q3 are located in a straight line. The position of Q2 is 0.279 m to the right of Q1. Q3 is located 0.180 m to the right of Q2. In the above problem, Q1 = 1.90 × 10-6 C, Q2 = -2.65 × 10-6 C, and Q3 = 3.18 × 10-6 C. Calculate the total force on Q2. Use a plus sign for a force directed to the right.


ok so first i try to calcuate the force of 1 on 2

so i use the equation f=(kq1q2)/r^2 ...so that would be
=(8.99E9*1.90E-6*-2.65E-6)/.279^2 = -.58150

or should i take the net electric charge and then calcuate the force of that charge at the point .279 away from q1?

(continue with first approch) then i calcuate the foce of 3 on 2
=(8.99E9*3.18E-6*-2.65E-6)/.180^2= -2.338

then i sum all the forces acting on 2 and i get -2.92 whic does not make sense.

what am i doing wrong

 

[Hootenanny]

(continue with first approch) then i calcuate the foce of 3 on 2
=(8.99E9*3.18E-6*-2.65E-6)/.180^2= -2.338

Why would this force be negative?

 

[Rasine]

well that's what i mean that my answere doesn't make sense...it turns out negtive because i am using the coulomb's law for the force of two charges and since it requires that i multiply the charges and one is negitive...then the answer is negtive

 

[Rasine]

should i calcuate the sum of the point charges for the system then use F=qoE to find the force where qo= the chrage of the object that i am trying to fined the foce about?

 

[Rasine]

if i do that...how do i calcuate the electric field about 2 because the distance is 0 so according to eqaution for point electic charge...the field would be 0 too

 

[Hootenanny]

No, I meant why you F₃₁ be negative, surely it will be positive since the force is acting to the right...

 

[Rasine]

that what i thought but in the formula i calls that i multiply q2 and q3 which is (-)(+)= - so that is why i am confused


what is wrong

 

[Hootenanny]

that what i thought but in the formula i calls that i multiply q2 and q3 which is (-)(+)= - so that is why i am confused


what is wrong

The negative sign indicates that the force will be attractive. However in this case we have defined a coordinate axis in which a rightwards acting force is considered positive;

Use a plus sign for a force directed to the right.

So rather than just computing the forces, you must consider the direction in which they act.

 

[Rasine]

the way that i was conidering it: i take q1 and q3 to be + because they are acting in the right direction...then q3 is - because is it actin to the left


so since this isn't wokring hoe can i take this into account better

 

[Hootenanny]

the way that i was conidering it: i take q1 and q3 to be + because they are acting in the right direction...then q3 is - because is it actin to the left

That's not quite correct. You need to consider the direction of the force acting on the charge q₂. So, q₁ is oppositely charged to q₂, therefore the force F₁₂ will be attractive and to the left, hence a negative answer (as you have got).

However, q₃ is again oppositely charge to q₂ and therefore the force F₃₂ will again be attractive, but this time the force is acting to the right. Therefore, the force should be positive.

Do you follow?

 

[Rasine]

oohhh ok. i will try again