- #1

WhiteWolf98

- 86

- 5

## Homework Statement

The magnitude of the electrostatic force between point charges ##q_1 = 26~\mu C## and ##q_2 = 47~\mu C## is initially ##F_1=5.70~N##. The separation distance between the charges, ##r_1## is then changed such that the magnitude of the force is, ##F_2=0.57~N##.

(a) What is the ratio of the new separation distance, ##r_2## to the initial separation, ##r_1##?

(b) What is the new separation distance, ##r_2##?

## Homework Equations

##F=K \cdot \frac {q_1 \cdot q_2} {r^2}##, where ##K \approx 8.99×10^9##

## The Attempt at a Solution

##5.70=\frac {K|q_1||q_2|} {{r_1}^2}##

##r_1=\sqrt {\frac {K|q_1||q_2|} {5.70}}##

##0.57=\frac {K|q_1||q_2|} {{r_2}^2}##

##r_2=\sqrt {\frac {K|q_1||q_2|} {0.57}}##

What's even the point of the ratio when all elements are present in the formula...?

##\frac {r_2} {r_1}=\frac {\sqrt {\frac {K|q_1||q_2|} {0.57}}} {\sqrt {\frac {K|q_1||q_2|} {5.70}}}##

##\frac {r_2} {r_1}=\frac {(\sqrt {\frac {K|q_1||q_2|} {0.57}})^2} {(\sqrt {\frac {K|q_1||q_2|} {5.70}})^2}##

##\frac {r_2} {r_1}=\frac {\frac {K|q_1||q_2|} {0.57}} {\frac {K|q_1||q_2|} {5.70}}##

##\frac {r_2} {r_1}=\frac {5.70} {0.57}##

##r_2=10r_1##

Working them out individually:

##K\cdot(47×10^{-6})(26×10^{-6})=10.98~ (to~3~s.f.)\gg (B)##

##5.70=\frac {B} {{r_1}^2}##

##r_1= \sqrt {\frac {B} {5.70}}=1.39~m~(to~3~s.f.)##

##r_2= \sqrt {\frac {B} {0.57}}=4.39~m~(to~3~s.f.)##

I'd expect ##r_2## to be larger, since the force is smaller. But it doesn't agree with the ratio... Have I made a wrong assumption or calculation somewhere?