Two metal spheres, each of mass 0.3kg, are oppositely charged by an electrostati

Question

Two metal spheres, each of mass 0.3kg, are oppositely charged by an electrostatic generator. Each is attached to an identical insulating string and suspended from the same point. Sphere A is negative with a charge magnitude three times that on sphere B. They attract eachother and hang with separation 0.4m. The length of the string from the point of support to the center of the ball is 2.5m.

1) Draw the electric field lines for this arrangement.
2) What is the electrostatic force on each sphere?
3) What is the charge on each sphere?
4) The two spheres are briefly connected by a copper wire. What is the final charge on each sphere?
5) What now is the separation distance between the two spheres?

Explanation / Answer

Charge on Sphere a = - 3q
Charge on Sphere b = q

Electrostatic Force, F = k*q*3q/r^2

As they have same mass and Same Force will be acting on each other, they wil make same anlge wrt Vertical.
Angle = sin^-1(0.3/2.5)
Angle = 6.9

Now, T*cos(6.9) = 0.3*9.8
T = 2.96 N

Electrostatic force on each sphere, = T*sin(6.9)
Electrostatic force on each sphere, = 2.96 * sin(6.9)
Electrostatic force on each sphere, = 0.355 N

k*q*3q/r^2 = 0.355
(9.0*10^9 * 3q^2)/0.4^2 = 0.355
q = 1.45 * 10^-6 C
q = 1.45 uC

Charge on Sphere A, = - 4.35 uC
Charge on Sphere B, = 1.35 uC

When they are connected together, Final Charge on each sphere, = (q-3q)/2 = -q
Final Charge on each sphere, = - 1.35 uC

Now they will Repel each other,
F = k*q^2/r^2

T*cos() = m*g
T*sin() =  k*q^2/r^2
tan() =  k*q^2/r^2/ m*g
(r-1)/2/sqrt( -((r-1)/2)^2 + 2.5^2) = (9.0*10^9 * (1.35*10^-6)^2)/(r^2*0.3*9.8)
r = 1.025 m

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