Physics, asked by khawajahasham, 4 months ago

Two light, conducting spheres, each 6mm diameter and having a mass of 10mg, are suspended
from the same point by fine insulating fibers 50cm long. Due to electrostatic repulsion, the
spheres are in equilibrium when 3cm apart. Calculate
a) the force between them.
b) the charge on each sphere.
c) combined potential at the surface of each sphere.

Answers

Answered by knjroopa

Explanation:

Given Two light, conducting spheres, each 6 mm diameter and having a mass of 10 mg, are suspended from the same point by fine insulating fibers 50 cm long. Due to electrostatic repulsion, the

spheres are in equilibrium when 3 cm apart. Calculate

a) the force between them.

b) the charge on each sphere.

c) combined potential at the surface of each sphere

If we consider a triangle with base 3 cm, opposite sides as 50 cm and T . Now draw a perpendicular in the middle of triangle. So we have 1.5 cm as opposite side and 50 cm to the hypotenuse.
Force F is in x direction and horizontal mg is in y direction
Now sine theta / 2 = 1.5 / 50
Or theta / 2 = sin^-1 1.5 / 50
= 1.719 degree
Now balanced forces in x direction will be
F = T cos (90 – 1.719)
T = F / cos 88.28
Now balanced forces in y direction will be
mg = T sin 88.28
T = mg / sin 88.28
Or F/mg = cot 88.28
Or F = mg cot 88.28
= 10 x 10^-3 x 9.8 cot 88.28
= 98 x 10^-3 x 0.03
= 2.9428 x 10^-3 N
Now the charges on both spheres can be got as
Electrostatic force F = kq^2 / r^2
q^2 = r^2 F / k
= (3 x 10^-2)^2 x 2.94 x 10^-3 / 9 x 10^9 (r = 3 cm apart and k is coulomb constant)
q^2 = 2.94 x 10^-7 x 10^-9
= 2.94 x 10^-16
Or q = 1.714 x 10^-8 C
Now combined potential will be
V = kq / r
= 9 x 10^9 x 1.714 / 3 x 10^-3 (r = d/2)
= 5.142 x 10^12 V

Reference link will be

https://brainly.in/question/14769651

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