#6**+15 **

Original force on charge 1 from the others f = K(q1*q2/d^{2} + q1*q3/(ā2*d)^{2} + q1*q4/d^{2}) 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.

.

Alan Feb 23, 2015

#3**+15 **

I believe the answer is C, they quadruple. Iām 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_

SevenUP Feb 21, 2015

#5**+5 **

**Alan**, could you please take a look at this question.

I think 315 and SevenUp are both keen to see your answer.

Melody Feb 22, 2015

#6**+15 **

Best Answer

Original force on charge 1 from the others f = K(q1*q2/d^{2} + q1*q3/(ā2*d)^{2} + q1*q4/d^{2}) 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.

.

Alan Feb 23, 2015