Q. 263.7( 3 Votes )
A particle having
Charge of the particle : q = 2.0 × 10–4 C
Distance between charged particle and the bob:
r =10 cm=0.1m
Mass of the bob : m = 100 g = 100 × 10-3 kg.
T = mg
Where T is the tension in the string and g is the acceleration due to gravity.
∴ T = 0.1 × 9.8
∴ T = 0.98 N
Now let the charge on the bob be : q’
Now the electrostatic force between the bob and the particle is given as:
This is the Coulomb’s Law. Where ϵ0 is the permittivity of free space and it’s value is : ϵ0 = 8.85418782 × 10-12 m-3 kg-1 s4 A2,q is the charge of the particle and q’ is the charge of the bob, r is the distance between charged particle and the bob.
Now, in order to have a loose string the tension in the string should be zero: T=0
For tension to be zero, the particle must repel the bob along the direction of the tension.
Also, weight of the bob is in opposite direction to the string.
Thus the equation is:
T + F = mg
Hence, the charge on the bob should be 5.4 × 10-4 C so that the string would become loose.
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