Electrostatics - Coulomb's Law

In summary, the ratio of the new separation distance, ##r_2## to the initial separation, ##r_1## is 1.39 to 3.
  • #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?
 
Physics news on Phys.org
  • #2
WhiteWolf98 said:

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.
 
  • Like
Likes WhiteWolf98
  • #3
Ah. I didn't forget, I didn't think you had to. Oops.

Would ##r_2=\sqrt 10 r_1## then?
 
  • #4
WhiteWolf98 said:
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.
 
  • Like
Likes WhiteWolf98
  • #5
I see. But even using that formula, it doesn't give me the same ##r_2## value
 
  • #6
WhiteWolf98 said:
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?
 
  • Like
Likes WhiteWolf98
  • #7
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
 

Related to Electrostatics - Coulomb's Law

1. What is Coulomb's Law?

Coulomb's Law is a fundamental law of electrostatics that describes the force between two electrically charged particles. It states that the force between two charged particles is directly proportional to the product of their charges and inversely proportional to the square of the distance between them.

2. What is the equation for Coulomb's Law?

The equation for Coulomb's Law is F = (k * q1 * q2) / r2, where F is the force between the two charged particles, k is the proportionality constant, q1 and q2 are the charges of the particles, and r is the distance between them.

3. How is Coulomb's Law related to Newton's Law of Universal Gravitation?

Coulomb's Law and Newton's Law of Universal Gravitation are similar in that they both describe the force between two objects. However, Coulomb's Law applies to electrostatic forces between charged particles, while Newton's Law of Universal Gravitation applies to the gravitational force between two objects with mass.

4. What are the units of measurement for the different variables in Coulomb's Law?

The unit of force (F) is measured in Newtons (N), the unit of charge (q) is measured in Coulombs (C), and the unit of distance (r) is measured in meters (m). The proportionality constant (k) has a unit of N*m2/C2.

5. How does the distance between two charged particles affect the force between them?

The force between two charged particles is inversely proportional to the square of the distance between them. This means that as the distance between the particles increases, the force between them decreases. Conversely, as the distance between the particles decreases, the force between them increases.

Similar threads

Replies
4
Views
881
  • Introductory Physics Homework Help
Replies
23
Views
468
  • Introductory Physics Homework Help
Replies
3
Views
1K
  • Introductory Physics Homework Help
Replies
6
Views
180
  • Introductory Physics Homework Help
Replies
3
Views
3K
  • Introductory Physics Homework Help
Replies
4
Views
724
  • Advanced Physics Homework Help
Replies
3
Views
673
  • Introductory Physics Homework Help
Replies
1
Views
759
  • Introductory Physics Homework Help
Replies
2
Views
1K
  • Introductory Physics Homework Help
Replies
4
Views
894
Back
Top