(adsbygoogle = window.adsbygoogle || []).push({}); 1. The problem statement, all variables and given/known data

Two identical positive point charges Q are nailed down a distance 2D apart. Point P is midway between the two charges as shwon in the diagram. A dashed line is vertical i.e. perpendiculat to an imaginary line connecting the two charges a distance D from both charges. Determine the magnitude and direction of the electric force acting on a positive charge Q_1 placed at the following locations.

a) at point p

b) on the dashed line a distance y above point p

c) where on the dashed line should Q_1 be placed so that if expierences the biggest force.

2. Relevant equations

3. The attempt at a solution

a) force at point p

welp i believe that both forces are pushing against eachother and because the point is in the center of the two charges that the two forces cancel out so

F = F-F = 0

b) force a distance y above point p

F = k(2Q)/r^2 where r^2 = D^2 + y^2

F = (1.38X10^-23)(2Q)/(D^2 + y^2) = (2.76X10^-23)Q/(D^2 + y^2)

c) find y so F is maximized so its beena while since i took calculus but i believe i have to take there derivitive of F interms of y, set it to 0, and check that its a maximum

also because F is a a quotient i remembered this rhym for the derivitive, Low D High minus High D Low over the square below.... which means you take the bottom times the derivitive of the top minus the top times the derivitive of the bottom, all over the bottom squared

F'(y) = ((D^2 + y^2)(0) - (2.76X10^-23)Q(2y))/((D^2 + y^2)^2) = -(2.76X10^-23)2Qy/((D^2 + y^2)^2

how can i set this to 0 with all the unknown variables.

it just occured to me that F is directly proportional to 1/r^2 so can i simply use that and ignore the other constants k & Q

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# Homework Help: Force on Point charges

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