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?
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
To find the magnitude of the force of interaction between B and C after the operations, we need to consider the principles of electrostatics.
Initially, when the system is in equilibrium with qA = 0 and qB = qC = q, the force between B and C is given by Coulomb's law:
F = k * (qB * qC) / r^2
where k is the electrostatic constant and r is the distance between the spheres.
Given that F = 4 N, we can rewrite the equation as:
4 = k * (q^2) / r^2 ...(1)
Now, let's analyze the effect of connecting the thin conducting wire between spheres A and B. When the wire is connected, the charges on A and B will start to redistribute, causing A to acquire a charge (let's say QA) and B to acquire a new charge (let's say QB).
Since the spheres are identical, the redistribution of charge will cause QA = QB = q/2. The sphere C remains unchanged with charge QC = q.
Next, when the wire is removed from A and B and connected between A and C, the charges redistribute again. This time, A acquires a new charge (let's say QA') and C acquires a new charge (let's say QC').
Since the charges redistribute evenly between A and C, both spheres will have the same final charge. Therefore, QA' = QC' = (QA + QC) / 2.
Now, let's calculate the final force between B and C. Using Coulomb's law again, we have:
F' = k * (QB * QC) / r^2 ...(2)
Substituting the new charges QA' = QC' and QB = QA = q/2 into equation (2), we get:
F' = k * [(q/2) * (QA + QC) / r^2]
Since QA = q/2 and QC = q, we can simplify the equation to:
F' = k * [(q/2) * (q/2 + q) / r^2]
= k * [(q/2) * (3q/2) / r^2]
= k * (3q^2 / 4r^2) ...(3)
From equations (1) and (3), we have the following ratio:
(4 / F) = (k * (q^2) / r^2) / (k * (3q^2 / 4r^2))
= (4r^2 / 3r^2) = 4/3
Therefore, the magnitude of the force of interaction between B and C after the operations is:
F' = (4 / F) * F
= (4 / 3) * 4
= 16/3 N
Hence, the magnitude of the force is 16/3 N.