Coulomb's law and negative charges

AI Thread Summary
The discussion centers on a confusion regarding a mathematical step in Coulomb's law involving negative charges. Participants clarify how the transition from (1.35keq2)/a2 to (1.91keq2)/a2 is achieved by squaring 1.35, multiplying by 2, and then taking the square root, resulting in approximately 1.91. One user expresses gratitude for the detailed explanation, emphasizing the importance of including all steps in mathematical texts. The conversation highlights the significance of clear mathematical reasoning in understanding physics problems. Overall, the clarification resolves the initial confusion about the calculation.
bobsmith76
Messages
336
Reaction score
0

Homework Statement



see attachment

Homework Equations





The Attempt at a Solution



Do you see the step where they go from (1.35keq2)/a2

to 1.91keq2)/a2 ?

I can't get that step. To my mind. If you square 1.35 then take its square root, you get 1.35 not 1.91
 

Attachments

  • Screen shot 2012-03-31 at 7.24.08 PM.png
    Screen shot 2012-03-31 at 7.24.08 PM.png
    6.8 KB · Views: 533
  • Screen shot 2012-03-31 at 7.24.13 PM.png
    Screen shot 2012-03-31 at 7.24.13 PM.png
    30.3 KB · Views: 615
Physics news on Phys.org
bobsmith76 said:

Homework Statement



see attachment

Homework Equations





The Attempt at a Solution



Do you see the step where they go from (1.35keq2)/a2

to 1.91keq2)/a2 ?

I can't get that step. To my mind. If you square 1.35 then take its square root, you get 1.35 not 1.91
\sqrt{1.35^2 \left (\frac{k_e q^2}{a^2} \right )^2 +1.35^2 \left (\frac{k_e q^2}{a^2} \right )^2}
\sqrt{\left (\frac{k_e q^2}{a^2} \right )^2(1.35^2 +1.35^2) }
\frac{k_e q^2}{a^2}\sqrt{1.35^2 +1.35^2 }
\frac{k_e q^2}{a^2}\sqrt{(2)(1.35^2)}
\frac{k_e q^2}{a^2}\sqrt{2}1.35
\frac{k_e q^2}{a^2}1.91
 
attachment.php?attachmentid=45754&d=1333236363.png


If you square 1.35 then multiply by 2, then take the square root of that result you do get 1.91, approximately.
 
Roshan,

Excellent answer. If only more math texts included as many steps as you do! I really appreciate it.
 
I multiplied the values first without the error limit. Got 19.38. rounded it off to 2 significant figures since the given data has 2 significant figures. So = 19. For error I used the above formula. It comes out about 1.48. Now my question is. Should I write the answer as 19±1.5 (rounding 1.48 to 2 significant figures) OR should I write it as 19±1. So in short, should the error have same number of significant figures as the mean value or should it have the same number of decimal places as...
Thread 'Collision of a bullet on a rod-string system: query'
In this question, I have a question. I am NOT trying to solve it, but it is just a conceptual question. Consider the point on the rod, which connects the string and the rod. My question: just before and after the collision, is ANGULAR momentum CONSERVED about this point? Lets call the point which connects the string and rod as P. Why am I asking this? : it is clear from the scenario that the point of concern, which connects the string and the rod, moves in a circular path due to the string...
Thread 'A cylinder connected to a hanging mass'
Let's declare that for the cylinder, mass = M = 10 kg Radius = R = 4 m For the wall and the floor, Friction coeff = ##\mu## = 0.5 For the hanging mass, mass = m = 11 kg First, we divide the force according to their respective plane (x and y thing, correct me if I'm wrong) and according to which, cylinder or the hanging mass, they're working on. Force on the hanging mass $$mg - T = ma$$ Force(Cylinder) on y $$N_f + f_w - Mg = 0$$ Force(Cylinder) on x $$T + f_f - N_w = Ma$$ There's also...
Back
Top