Question

A small spherical insulator of mass 5.49 × 10-2 kg and charge +0.600 μC is hung by a thin wire of negligible mass. A charge of -0.900 μC is held 0.150 m away from the sphere and directly to the right of it, so the wire makes an angle with the vertical (see the drawing). Find (a) the angle and (b) the tension in the wire.

Answer 1

Given:

Mass of the spherical insulator = 5.49 ×10^-2 kg

Charge on the spherical insulator = +0.600 μC = 0.6 ×10^-6 C = +6 ×10^-7 C

Charge held right of insulator q = -0.900 μC = 0.9 ×10^-6 C = - 9 ×10^-7 C

Distance between spherical insulator and charge  r = 0.15 m

In given case forces acting on the sphere are,

1) weight on the sphere mg in downward direction.

2) Electrostatic attraction between spherical insulator and charge F_E.   

_{According to Coulombs law}   

3) Tension in the wire T.

(1)

In this case the spherical insulator is in equilibrium that means all forces balance each other therefore Net force on the spherical insulator is zero.

now resolve the tension in the wire into two component .

Vertical component is   and

horizontal component is  

Balancing the horizontal forces on the insulator.

Balancing the vertical forces on the insulator.

Dividing equation (1) by equation (2)

Substituting all the values,

(2)

putting the value of angle in equation (2) we get,

therefore angle is 21.8⁰ and Tension is 0.580 N.