Homework Help: Force on Point charges

1. Apr 26, 2010

joemama69

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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2. Apr 26, 2010

thebigstar25

part (a) in the question is correct ..

for part (b) and (c) I suggest you take another approach ..

you know that F = q*E (where q is the charge, and E is the electric field) ..

hint: before you find the force, find the electric field acting at the points for part(b) and (c).. starting with part(b) take in consideration that the electric field is a vector and it can be expressed in x and y components .. things will get easier this way ..

3. Apr 26, 2010

joemama69

so E will then be the sum of the two fields produced by the two positive charges Q

E = kQ $$\vec{r_1}$$/r2 + kQ $$\vec{r_2}$$/r2 = (kQ/r2) [(D$$\vec{x}$$ + y$$\vec{j}$$) + (-D$$\vec{x}$$ + y$$\vec{j}$$)) = (kQ/r2) (2y$$\vec{j}$$)

Last edited: Apr 26, 2010
4. Apr 26, 2010

thebigstar25

good, the x component cancels .. Go on :)

5. Apr 26, 2010

joemama69

so do i just leave my answer like that and plug it into the force equatin

F = E/q = (kQ/r2)(2yj)

so to find the max i integrate F interms of y, shuld i put r2 = D2 + y2

6. Apr 26, 2010

thebigstar25

would you please explain why the "y" there .. why didnt you write E as Ex i + Ey j which is equal to Ecos(theta) i + Esin(theta) j ..

7. Apr 27, 2010

joemama69

because the y is the vertical distance it is away from the other charges... it looks like you went a step further and denoted it as y = sin(theta)

So is the force then... F = 2Esin(theta) j = 2kQsin(theta)/r2 and i will differentiate interms of theta. do i have to change the r^2 into an element of theta as well

Last edited: Apr 27, 2010
8. Apr 27, 2010

thebigstar25

that would be true if the two points are exactly at the point p, but in part b you can not use y to be the distance between the charge and the point where you want to find the electric field ..

thats what you have to use ..

http://img696.imageshack.us/img696/6738/62871910.jpg [Broken]

if you disagree with that, please say why ..

Last edited by a moderator: May 4, 2017