Three identical conducting spheres are located at the vertices of an equilateral triangle ABC. Initially the charge the charge of the sphere at point A is qA=0and the spheres at B and C carry the same charge qB=qC=q. It is known that the sphere B exerts an electrostatic force on C which has a magnitude F=4 N. Suppose an engineer connects a very thin conducting wire between spheres A and B. Then she removes the wire and connects it between spheres A and C. After these operations, what is the magnitude of the force of interaction in Newtons between B and C?

asked by stranger

11 years ago

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F(BC) =kq²/a²
q´(B)=q´(A) =(q+0)/2=q/2,
q´´(A) =q´(C) = {q´(A)+c(C)}/2 = 3q/4
F´(BC) =k q´(B)• q´(C)/a² =
=(3/8)• kq²/ a²=3•4/8=3.2 = 1.5 N

answered by Elena

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Three identical conducting spheres are located at the vertices of an equilateral triangle ABC. Initially the charge the charge of the sphere at point A is qA=0and the spheres at B and C carry the same charge qB=qC=q. It is known that the sphere B exerts an electrostatic force on C which has a magnitude F=4 N. Suppose an engineer connects a very thin conducting wire between spheres A and B. Then she removes the wire and connects it between spheres A and C. After these operations, what is the magnitude of the force of interaction in Newtons between B and C?

1 answer

F(BC) =kq²/a²
q´(B)=q´(A) =(q+0)/2=q/2,
q´´(A) =q´(C) = {q´(A)+c(C)}/2 = 3q/4
F´(BC) =k q´(B)• q´(C)/a² =
=(3/8)• kq²/ a²=3•4/8=3.2 = 1.5 N

Elena · Answer

F(BC) =kq²/a² q´(B)=q´(A) =(q+0)/2=q/2, q´´(A) =q´(C) = {q´(A)+c(C)}/2 = 3q/4 F´(BC) =k q´(B)• q´(C)/a² = =(3/8)• kq²/ a²=3•4/8=3.2 = 1.5 N