# Calculating total force on charged particle

## Homework Statement

A particle with charge 3 μC is located on the x-axis at the point -10 cm, and a second particle with the charge 4 μC is placed on the x-axis at -4 cm.

What is the magnitude of the total electrostatic force on the third particle with the charge 7 μC placed on the x-axis at -2 cm? The Coulomb constant is 8.9875e9.

## Homework Equations

$$F=(k*Q_1*Q_2)/(r^2)$$

## The Attempt at a Solution

I realize how simple this problem is but the stupid online software keeps telling me I'm wrong (which I very well might be). I converted μC to C and cm to m. I keep getting 652, so with sig figs, 700 N.

What am I doing wrong? Thanks!

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gneill
Mentor
Can you show more of your work? What magnitudes and directions are you getting for the individual forces?

This is what I am doing.

$ƩF=8.9875*10^9(\frac{.000003*.000007}{.08^2}+\frac{.000004*.000007}{.02^2})$

I assume its all in the same direction considering that's all the info they gave me in the question.

gneill
Mentor
The forces won't be in the same direction! All the charges are positive and the third charge is situated BETWEEN the other two. Draw a diagram...pencil in the force vectors...

Check your value for the separation of the 3 and 7 μC charges.

Oh crap, it is supposed to be -10 cm not -.10 cm. So so sorry.

In this case, they are in the same direction, yes?

gneill
Mentor
Oh crap, it is supposed to be -10 cm not -.10 cm. So so sorry.

In this case, they are in the same direction, yes?

Ah. Then yes, the forces will act in the same direction.

I'm seeing a net value a bit higher than what you've calculated (affects the rounding). Try keeping more digits in your intermediate calculations. The number of significant figures required is a bit of a puzzle, since all the values except for the constant k seem to be given as exact figures (no decimal points indicated). Rounding to one sig fig could be an issue, I'm not sure. Be sure to specify the units on the result, too, since most automated systems are picky that way.

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Ok, thanks a bunch gneill!