Two pitch balls are charged by rubbing against each other. They are suspended from a horizontal rod through two strings of length 20cm each, the separation b/w the suspension point being 5cm. In equilibrium, the separation b/w balls is 3cm. Find the mass of each ball & tension in string.
The charge on each ball has a magnitude 2X10-8c
11
q = 2.0 × 10–8 c
T = ? Sin theta = 1/20
Force between the charges
Fï€=k q1 q2 / r 2 = k q2 / r2
=(9*10 9 )(2*10 -8 )2 / (3) 2
= 4 × 10–3 N
mg sin theta =F , therefore,
m = F / g sin theta
= (4*10-3) / (10) (1/20) = 8 gms
Since cos2 theta + sin 2theta = 1 , therefore,
cos theta = (1- sin2 theta) 1/2
= √(399/400)
=0.99
So, T = mg sin theta , we get
T= 8 * 10 -3 * 0.99 = 8 *10 -2 .........Ans
1dat one is easy..just balance the horizontal n vertical component of electrostatic force n mg........n tension..
1The method is only to get the Equation is
Tsinθ=Q24Πε0 r2
and Tcosθ= mg Just Solve these two Equations to get the answer
