What Is the Ratio of the New Separation Distance to the Initial Separation?

[WhiteWolf98]

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?

 

[PeroK]

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...?

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

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

>>

The above step is an error. Did you forget to square the left hand side?

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

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

$r_2=10r_1$

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?

See above.

 

[WhiteWolf98]

Ah. I didn't forget, I didn't think you had to. Oops.

Would $r_2=\sqrt 10 r_1$ then?

 

[PeroK]

Ah. I didn't forget, I didn't think you had to. Oops.

Would $r_2=\sqrt 10 r_1$ then?

Yes. It's an inverse square law. If the force reduces by a factor the distance increases by the square root of that factor.

 

[WhiteWolf98]

I see. But even using that formula, it doesn't give me the same $r_2$ value

 

[haruspex]

I see. But even using that formula, it doesn't give me the same $r_2$ value

It's as close as can be expected after rounding the individual distances to three sig figs.
What are you getting for the ratio? What if you take an extra digit in the rounding?

 

[WhiteWolf98]

Okay, for some reason I did it again, and it worked... either way, not complaining. Thank you both. How silly for it all to be just an algebra mistake in the end