Question

22. Two balls of same radius and mass are suspended on
threads of length 1 m as shown. The mass of each ball
and charge is 15 g and 126 HC respectively. When the
balls are in equilibrium, the separation between them is
15 g
0 - 126 CK
8 cm
15 9
Q = 126 C
8 cm. The new saparation between them when one of the
balls is discharged to half of original charge, is:
Nar 5 cm
(b) 6 cm
ccm
(d) 2 cm​

Answer 1

Two balls of same radius and mass are suspended on the threads of length 1m. The mass of each ball and charge is 15g and 126 C respectively. the seperation between them is 8cm.

To find : The new seperation between them when one of the balls is discharged to half of original charge.

solution : let new seperation between them is x.

let tension in string is T and angle between threads is 2θ

at equilibrium,

Tcosθ = mg ......(1)

Tsinθ = kq₁q₂/x² ........(2)

from equations (1) and (2) we get,

tanθ = kq₁q₂/x²/mg

⇒(x/2)√{l² - (x/2)²} = kq₁q₂/mgx²

as l << x , so √{l² - x²/4} ≈ l

⇒x/2l = kq₁q₂/mgx²

⇒x³ = 2lkq₁q₂/mg

if l and m are constant then, x³ ∝ q₁q₂

i.e., (x/x')³ = (q₂/q₂')

initial seperation between them, x = 8 cm

initial charge , q₁q₂ = 126 C × 126 C and final charge, q₁'q₂' = 63 C × 63 C

so, (8/x')³ = (126 × 126/63 × 63)

⇒8³/x³ = 4

⇒x = 8/(4)⅔ ≈ 5 cm

Therefore the new seperation between them is 5 cm

Answer 2

Answer:

5cm

Explanation:

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