HCV part 2 Pg. 121 q.22 

•     2 identical balls each having a chrg of 2 x 10^ -7 and a mass of 100g , r suspended from a common pt. by 2 insulationg strips each 50 cm long.. The balls are held at a separation 5 cm apart & then released . (a0 Find the elecric force on one of the carged balls.  (b) the components of the resultant force on it along and perpendicular to the string. (c) the tension in the sting . (c) the acceleration of one of the balls  . Answers are to be obtaioned only for the instant just ahter the release.

22) q₁ = q₂ = 2 × 10^–7 c  ,  m = 100 g

l = 50 cm = 5 × 10^–2 m  ,  d = 5 × 10^–2 m

(a) Now Electric force

F = Kq²/r²  implies  9*10⁹*4*10 ^-14 / 25*10 ^-4  N  implies 14.4 N = 0.144N

(b) The components of Resultant force along it is zero,

because mg balances T cos$ and so also.

F = mg = T sin$

(c) Tension on the string

T sin$ = F  , T cos$ = mg   implies  T² = F²+(mg)²  implies (0.144)² + (0.1*9.8)²  implies T = 0.986 N

Tan$ = F/mg  =  0.144 / 100*10 ^-3*9.8  =  = 0.14693

But T cos$= 10² × 10^–3 × 10 = 1 N

implies T = 1/cos$  = sec$

implies T = F/sin$

Sin$ = 0.145369 ; Cos$ = 0.989378