Coulomb's law and negative charges

[bobsmith76]

Homework Statement

see attachment

Homework Equations

The Attempt at a Solution

Do you see the step where they go from (1.35k_eq²)/a²

to 1.91k_eq²)/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

 

[RoshanBBQ]

Homework Statement

see attachment

Homework Equations

The Attempt at a Solution

Do you see the step where they go from (1.35k_eq²)/a²

to 1.91k_eq²)/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

$$\sqrt{1.35^2 \left (\frac{k_e q^2}{a^2} \right )^2 +1.35^2 \left (\frac{k_e q^2}{a^2} \right )^2}$$
$$\sqrt{\left (\frac{k_e q^2}{a^2} \right )^2(1.35^2 +1.35^2) }$$
$$\frac{k_e q^2}{a^2}\sqrt{1.35^2 +1.35^2 }$$
$$\frac{k_e q^2}{a^2}\sqrt{(2)(1.35^2)}$$
$$\frac{k_e q^2}{a^2}\sqrt{2}1.35$$
$$\frac{k_e q^2}{a^2}1.91$$

 

[SammyS]

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

 

[bobsmith76]

Roshan,

Excellent answer. If only more math texts included as many steps as you do! I really appreciate it.