- #1

kavamo

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**Coulombs Law--what now?**

## Homework Statement

Three point charges are fixed in place in a right triangle. What is the electric force on the q = -0.64 µC charge due to the other two charges? (Let Q1 = +0.71 µC and Q2 = +1.1 µC.)

Find:

magnitude N

and

direction ° above the positive x-axis

## Homework Equations

F=kq1q2 / r^2

K=8.99 x 10^9

## The Attempt at a Solution

The diagram from the book shows a right triangle where q is the 90 degrees vertex along X-axis. the hypotenuse = 10 cm; r (y)=8.0 cm; and between q & Q2 =6 cm.

I have made free body diagrams and know: q = -0.64 µC; on the y-axis Q1= +0.71 µC and Q2 = +1.1 µC on the X-axis

plugging the info into the formulas I get:

F (y)= (8.99x10^9)(0.71)(0.64) / 8^2 = 6382900

F (x)= (8.99x10^9)(1.1)(0.64) / 6^2 = 175804444.4

Don't these numbers seem rather big? Did I miss a conversion somewhere?

I am also unsure of what to do next.

Thanks in advance for your help.