Coulomb's law and negative charges

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Homework Help Overview

The discussion revolves around the application of Coulomb's law in the context of negative charges, specifically focusing on a mathematical step involving the transition from one expression to another in a calculation related to electric forces.

Discussion Character

  • Mathematical reasoning

Approaches and Questions Raised

  • Participants are examining a specific mathematical step where the transition from (1.35keq²)/a² to (1.91keq²)/a² is questioned. There is confusion regarding the calculation involving squaring and taking square roots of the number 1.35.

Discussion Status

Some participants have provided clarifications regarding the mathematical operations involved, noting that squaring 1.35 and multiplying by 2 before taking the square root leads to the approximate value of 1.91. There is acknowledgment of the clarity in the explanations provided by one participant.

Contextual Notes

Participants are working from an attachment that contains the original problem statement and equations, which are not fully visible in the discussion. There is an emphasis on the steps taken in the calculations rather than the final outcome.

bobsmith76
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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
 

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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.
 

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