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Hello, I have a simple exercice I'm having trouble to solve.

So we have a) and b)

a) asked the force applied on a charge.... and the answer was [tex] \vec{F_{res}.}= Qkq ( -0,222 \vec{i} -0,250\vec{j}) N [/tex]

that's not a problem here.. that was the good answer. Now in b), they're asking where do they have to put a +2,5Q charge to nullify the force exerced.. so I thought yeah really simple, the force just has to be the inverse:

[tex] \vec{F_{12}}= Qkq ( 0,222 \vec{i} + 0,250\vec{j}) N [/tex]

and since [tex] \vec{F_{12}}= \frac{2,5Qkq}{r^2_{21}} \vec{u_{21}} [/tex]

then I just have to equal this last equation to the 2nd to last one, and I equal the [tex] \vec{i} [/tex] and [tex] \vec{j} [/tex] vectors in both ones to find out the vector of where the charge should be placed to nullify the force applied on it, but I'm having trouble finding it. Believe me, I have tried more than 10 times, and I can't find the vector. Can someone please help me on this one?

thanks a lot!

So we have a) and b)

a) asked the force applied on a charge.... and the answer was [tex] \vec{F_{res}.}= Qkq ( -0,222 \vec{i} -0,250\vec{j}) N [/tex]

that's not a problem here.. that was the good answer. Now in b), they're asking where do they have to put a +2,5Q charge to nullify the force exerced.. so I thought yeah really simple, the force just has to be the inverse:

[tex] \vec{F_{12}}= Qkq ( 0,222 \vec{i} + 0,250\vec{j}) N [/tex]

and since [tex] \vec{F_{12}}= \frac{2,5Qkq}{r^2_{21}} \vec{u_{21}} [/tex]

then I just have to equal this last equation to the 2nd to last one, and I equal the [tex] \vec{i} [/tex] and [tex] \vec{j} [/tex] vectors in both ones to find out the vector of where the charge should be placed to nullify the force applied on it, but I'm having trouble finding it. Believe me, I have tried more than 10 times, and I can't find the vector. Can someone please help me on this one?

thanks a lot!

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