- #1

logearav

- 338

- 0

## Homework Statement

The sum of two point charges is 6µC. They attract each other with a force of 0.9 N, when kept 40 cm apart in vacuum. Calculate the charges.

## Homework Equations

## The Attempt at a Solution

q

_{1}+(-q

_{2}) = 6x10

^{-6}

q

_{1}-q

_{2}=6x10

^{-6}...I

q

_{1}=6x10

^{-6}+q

_{2}

According to Coulomb's law

F=q

_{1}q

_{2}/4πε

_{0}r

^{2}

.9=(6x10

^{-6}+q

_{2})q

_{2}/4πε

_{0}(.4)

^{2}

.9=(6x10

^{-6}q

_{2}+q

_{2}

^{2})x9x10

^{9}/.16

.144/9x10

^{9}=6x10

^{-6}q

_{2}+q2

^{2}

144x10

^{-3}x10

^{-9}/9=6x10

^{-6}q

_{2}+q

_{2}

^{2}

q

_{2}

^{2}+6x10

^{-6}q

_{2-}16x10

^{-12}=0

(q

_{2}+8x10

^{-6})(q

_{2}-6x10

^{-6})=0

q

_{2=}-8µC or q

_{2}=6µC

The answers given in the book are q

_{1}= 8micro Coulomb and q

_{2}= -2 microcoulomb

I don't understand how to get this answer because when q

_{2}= 8 microcoulomb and substituting in equation I , i get q

_{1}+8micro coulomb =6 micro coulomb so q

_{1}should be -2 micro coulomb

Did i miss anything in the steps? Members, please help me .