# Electric charge

1. May 11, 2007

### 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

2. May 11, 2007

### Hootenanny

Staff Emeritus
Why would this force be negative?

3. May 11, 2007

### Rasine

well thats what i mean that my answere doesn't make sense...it turns out negtive becuase 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

4. May 11, 2007

### 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?

5. May 11, 2007

### Rasine

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

6. May 11, 2007

### Hootenanny

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

7. May 11, 2007

### 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

8. May 11, 2007

### Hootenanny

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

9. May 11, 2007

### Rasine

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

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

10. May 11, 2007

### Hootenanny

Staff Emeritus
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?

11. May 11, 2007

### Rasine

oohhh ok. i will try again