+33615 Original force on charge 1 from the others f = K(q1*q2/d2 + q1*q3/(√2*d)2 + q1*q4/d2) where K is a constant (I've assumed d1 and d3 are on opposite corners).
(Strictly, we should add these as vectors, but that's an unnecessary complication here.)
If all the q terms are multiplied by 2 and everything else remains the same it is easy to see that a factor of 4 increases each term, and hence the force on charge 1 quadruples. The same argument applies to all the other charges, so SevenUp is correct.
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+394 I believe the answer is C, they quadruple. I’m not absolutely certain, but it is logical because each charge increases 100% and there are 4 charges. I know that the potential at any point on the square is the algebraic sums of the individual charges. So a change of magnitude of a charge is directly proportional.
Alan can verify the accuracy or error of my statements.
Your question was lost in the sea, even after Melody bumped it.
_7UP_
+118609 Alan, could you please take a look at this question. 
I think 315 and SevenUp are both keen to see your answer. 
+33615 Original force on charge 1 from the others f = K(q1*q2/d2 + q1*q3/(√2*d)2 + q1*q4/d2) where K is a constant (I've assumed d1 and d3 are on opposite corners).
(Strictly, we should add these as vectors, but that's an unnecessary complication here.)
If all the q terms are multiplied by 2 and everything else remains the same it is easy to see that a factor of 4 increases each term, and hence the force on charge 1 quadruples. The same argument applies to all the other charges, so SevenUp is correct.
.
