Electris Field Quest confuzzled

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SUMMARY

The discussion centers on determining the point between two point charges, A (3.5 x 10^-6 C) and B (2.5 x 10^-6 C), where the electric field strength is zero, separated by 15 cm. The solution involves setting the electric fields from both charges equal (Ea = Eb) and solving for the distance 's' from charge A. The critical mathematical step involves expanding the equation from Step 2 into a standard quadratic form, leading to the equation 0.3s² - 30s + 225 = 0, which is solved to find that 's' equals 8.1 cm from charge A.

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Homework Statement



Two point chagres A= 3.5 x 10^-6 and B= 2.5 x 10^-6 are separated by a distance of 15 cm. Determine the point between the charges at which the electric field strength is zero.

this was an example in the physics textbook i use and this is the answer they came up with


Ea=Eb

Step 1: 9.00 x 10^9 x 3.5 x 10^-6/ s^2 = 9.00 x 10^9 x 2.5 x 10^-6/ (15 - s)^2

Step 2: 3.5 x 10^-6 (15-s)^2= 2.5 x 10^-6 x s^2
Step 3: 0.3s^2 - 30s + 225= 0
s= 8.1 cm from A

I understand it all until Step 3: what have they done there?
 
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They expanded step 2 and set it up so that s could be solved as a root value (its a mathematical step, not much physical analysis).
 

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