Neutral metal sphere A, of mass 0.10 kg, hangs from an insulating wire 2.0 m long. An identical metal sphere B, with charge 2q, is brought into contact with sphere A. The spheres repel and settle

Question

Calculate the initial charge on B.

|\ <--- The angle inside the triangle is 12 degrees.

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B----- A

Right angle triangle

Answer 1

From the look of the drawing, it appears that metal sphere B stays in place while the metal sphere A hanging from the wire is repelled from the first at an angle of 12 degrees.

There are 3 forces acting on sphere A: the force of gravity mg, the repulsive electrical force from sphere B, and the y component of the tension force from the wire.

When the spheres contact, the charge 2Q is spread out to both spheres, so each sphere has a charge Q.

the repulsive force on sphere A is k Q² / d² where d is the distance separating the spheres.

d is given by 2.0 sin12 = 0.42 m.

The gravitation force on sphere B is 0.1 kg x 9.8 ms^-1 = 0.98 N.

This is balanced by the downward component of the tension force = T x cos 12,

so T x cos12 = 0.98 N or T = 1.0 N .

The repulsive force on B is balanced by the horizontal component of the tension force which is

1.0 N x sin 12 = 0.21 N

Finally, equate the repulsive force to the horizontal component of the tension force and solve for Q, then multiply by 2.

0.21 N= 8.99 x 10⁹ x Q² / 0.42² m² . Q = 2.0 x 10^-6 C and 2Q = 4.0 x 10^-6C.

Please check the math.