# In the vertex of an equilateral triangle are set 3 identical loads, q. What load has to be set in the center of triangle, for obtaining equilibrium?Is the equilibrium stable? Calculate the...

In the vertex of an equilateral triangle are set 3 identical loads, q. What load has to be set in the center of triangle, for obtaining equilibrium?

Is the equilibrium stable? Calculate the potential energy of the system.

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A loaded particle, set in an electrical field, can stay in repause if only in the place where it is set, the electrical field is nule.

We'll apply the principle of superposition:

The resultant field in a specific point is equal to the vectorial sum of the fields produced separately and independently by the each and every load itself.

In the place where the load Q is set, the resultant field, produced by the 3 loads q, is nule, based on symmetry principle. We'll have to calculate the filed from the triangle's vertex and to put the annulement condition.

We'll have 3 contributions E1=E2 and E3 and the load Q have to have an opposite orientation to q. If all loads would have the same sign, the system will desintegrate immediately.

2E1*cos 30-E3=0

E1=q/(4*pi*epsilon*l^2)

E3=module Q/(4*pi*epsilon*(l/sqrt 3)^2)2[q/(4*pi*epsilon*l^2)](sqrt 3)/2==module Q/(4*pi*epsilon*(l/sqrt 3)^2)module Q=q/sqrt 3The condition above doesn't depend on l, so the system stays in equilibrium for any l, if the equilateral triangle configuration is kept, otherwise the equilibrium is instable.The potential energy of the system:Ep=1/2 sum (qi*qk)/(4*pi*epsilon*r ik)=sum 1/2 qi*ViVi is the field's potential, in the point where is placed qiEp=1/2*3q(2q/(4*pi*epsilon*l)-(q/sqrt3)/(q/(4*pi*epsilon*l)/sqrt3))+1/2*(-q/sqrt 3)*3q/q/(4*pi*epsilon*sqrt 3)=0So in the given configuration, the potential energy stays nule.**Ep=0**