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Two equally charged pith balls each of mass 10gm are suspended from the same point by two silk thread each of length 1.2m. As a result of mutal repulsion the balls are separated by 5cm. Find the charge on each ball
Asked by bhuvanshetty1340 | 17 Apr, 2020, 05:56: PM
Expert Answer
Figure shows the two pith balls of charge of equal magnitude q positioned due to repulsive force.
Tension forces in the string are resolved into horizontal and vertical components.
Vertical component Tcosθ balances the weight mg. Horizontal component Tsinθ balances the electrostatic force qE
T sinθ = Fe = q2 / [ (4πεo) d2 ] .......................( 1 )
where d = 5 cm is the separation distance between charges
T cosθ = mg ......................(2)
By dividing eqn.(1) with eqn.(2) , tanθ = q2 / [ mg (4πεo) d2 ]
we get charge q from above equation as, q = { [ mg (4πεo) d2 ] tanθ }1/2 ...................... (1)
where tanθ = 2.5/ { [ 1202 - ( 2.5*2.5) ]1/2 } ≈ 2.5/120 = 0.021
q = { [ 10 × 10-3 × 9.8 × ( 10-9 / 9 ) × 25 × 10-4 ] × 0.021 }1/2 = 2.4 × 10-9 C
Answered by Thiyagarajan K | 17 Apr, 2020, 09:48: PM
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